CBSE Class 10 Maths Important Questions - Chapter 6 Triangles

1 Mark Question

1. In the figure $$△\mathbf{ABC}\sim △\mathbf{EDC}\mathbf{,}$$if we have $$\mathbf{\text{AB}}\mathbf{=}\mathbf{4}\mathbf{\text{cm}}\mathbf{,}\mathbf{\text{ED}}\mathbf{=}\mathbf{3}\mathbf{\text{cm}}\mathbf{,}\mathbf{\text{CE}}\mathbf{=}\mathbf{4.2}\mathbf{\text{cm}}$$and $$\mathbf{\text{CD}}\mathbf{=}\mathbf{4.8}\mathbf{\text{cm}}\mathbf{,}$$then the values of $$\mathbf{\text{CA}}\mathbf{and}\mathbf{\text{CB}}$$ are

1. $$\mathbf{6}\mathbf{\text{cm}}\mathbf{,}\mathbf{6.4}\mathbf{\text{cm}}$$

2. $$\mathbf{4.8}\mathbf{\text{cm}}\mathbf{,}\mathbf{6.4}\mathbf{\text{cm}}$$

3. $$\mathbf{5.4}\mathbf{\text{cm}}\mathbf{,}\mathbf{6.4}\mathbf{\text{cm}}$$

4. $$\mathbf{5.6}\mathbf{\text{cm}}\mathbf{,}\mathbf{6.4}\mathbf{\text{cm}}$$

Ans: (d)  $5.6\text{cm},6.4\text{cm}$ 

As the triangles are similar,

 ${\displaystyle \frac{AB}{ED}}={\displaystyle \frac{CA}{CE}}={\displaystyle \frac{CB}{CD}}={\displaystyle \frac{4}{3}}$ 

 ${\displaystyle \frac{CA}{CE}}={\displaystyle \frac{CA}{4.2}}={\displaystyle \frac{4}{3}}$ 

 $CA=5.6$ 

Then,

 ${\displaystyle \frac{CB}{CD}}={\displaystyle \frac{CB}{4.8}}={\displaystyle \frac{4}{3}}$ 

 $CB=6.4$ 

2. The areas of two similar triangles are respectively  $\mathbf{9}\mathbf{\text{c}}{\mathbf{\text{m}}}^{\mathbf{2}}$ and $\mathbf{16}\mathbf{\text{c}}{\mathbf{\text{m}}}^{\mathbf{2}}$ . Then ratio of the corresponding sides are

1. 3:4  

2. 4:3  

3. 2:3 

4. 4:5 

Ans: (a)  $3:4$ 

Ratio of area of triangle is equal to the square the ratio of the corresponding sides

Therefore,  $\sqrt{{\displaystyle \frac{9}{16}}}={\displaystyle \frac{3}{4}}$ 

3. Two isosceles triangles have equal angles and their areas are in the ratio $$\mathbf{16}\mathbf{:}\mathbf{25}\mathbf{,}$$then the ratio of their corresponding heights is

1. ${\displaystyle \frac{\mathbf{4}}{\mathbf{5}}}$ 

2. ${\displaystyle \frac{\mathbf{5}}{\mathbf{4}}}$ 

3. ${\displaystyle \frac{\mathbf{3}}{\mathbf{6}}}$ 

4. ${\displaystyle \frac{\mathbf{5}}{\mathbf{7}}}$ 

Ans: (a)  ${\displaystyle \frac{4}{5}}$ 

Ratio of area of triangle is equal to the square the ratio of the corresponding sides

Therefore,  $\sqrt{{\displaystyle \frac{16}{25}}}={\displaystyle \frac{4}{5}}$ 

4. If  $△\mathbf{ABC}\sim △\mathbf{DEF}$ and  $\mathbf{AB}\mathbf{=}\mathbf{5}\mathbf{\text{cm}}\mathbf{,}$ area  $\mathbf{(}\mathbf{\Delta }\mathbf{ABC}\mathbf{)}\mathbf{=}\mathbf{20}\mathbf{\text{c}}{\mathbf{\text{m}}}^{\mathbf{2}}\mathbf{,}$ area  $\mathbf{(}\mathbf{\Delta }\mathbf{DEF}\mathbf{)}\mathbf{=}\mathbf{45}\mathbf{\text{c}}{\mathbf{\text{m}}}^{\mathbf{2}}\mathbf{,}$ then  $\mathbf{\text{DE}}\mathbf{=}$ 

1. ${\displaystyle \frac{\mathbf{4}}{\mathbf{5}}}\mathbf{\text{cm}}$ 

2. $\mathbf{7.5}\mathbf{\text{cm}}$ 

3. $\mathbf{8.5}\mathbf{\text{cm}}$ 

4. $\mathbf{7.2}\mathbf{\text{cm}}$ 

Ans: (b)  $7.5\text{cm}$ 

Ratio of area of triangle is equal to the square the ratio of the corresponding sides

Therefore,  $\sqrt{{\displaystyle \frac{(\mathrm{\Delta }ABC)}{(\mathrm{\Delta }DEF)}}}={\displaystyle \frac{AB}{DE}}$ 

 $\sqrt{{\displaystyle \frac{20}{45}}}={\displaystyle \frac{5}{DE}}\Rightarrow DE=7.5$ 

 ${\displaystyle \frac{(\mathrm{\Delta }ABC)}{(\mathrm{\Delta }DEF)}}={\displaystyle \frac{20}{45}}={\displaystyle \frac{5}{7.5}}$ 

5. A man goes  $15\text{m}$ due west and then  $8\text{m}$ due north. Find distance from the starting point.

1. 17 

2. 18

3. 16

4. 7

Ans: (A)  $17\text{m}$ 

Square of the distance = square of the sum of two distances.

 $=\sqrt{{15}^2+8^2}=\sqrt{225+64}=\sqrt{289}=17m$ 

6. In a triangle  $\text{ABC},$ if  $\text{AB}=12\text{cm},\text{BC}=16\text{cm},\text{CA}=20\text{cm},$ then  $△ABC$ is

1. Acute angled

2. Right angled

3. Isosceles triangle

4. Equilateral triangle

Ans: (b) Right angled

By law of Pythagoras triangle,

 $A{C}^2=B{C}^2+A{B}^2={\left(16\right)}^2+{\left(12\right)}^2=400$ 

 $AC=20$ 

7. In an isosceles triangle  $\text{ABC},\text{AB}=\text{AC}=25\text{cm}$ and  $\text{BC}=14\text{cm},$ then altitude from  $\text{A}$ on  $\text{BC}=$ 

1. 20

2.  24

3. 12

4. None of these

Ans: (b)  $24\text{cm}$ 

8. The side of square who's diagonal is  $16\text{cm}$ is

1. $\mathbf{16}\mathbf{\text{cm}}$ 

2. $\mathbf{8}\sqrt{\mathbf{2}}\mathbf{\text{cm}}$ 

3. $\mathbf{5}\sqrt{\mathbf{2}}\mathbf{\text{cm}}$ 

4. None of these

Ans: (a) $$8\sqrt{2}\text{cm}$$

Length of diagonal = sum of both sides of square.

 ${\left(16\right)}^2=s^2+s^2$ 

 $2s^2=16\ast 16$ 

 $s^2=8\left(16\right)$ 

 $s=8\sqrt{2}$ 

9.In an isosceles triangle  $\text{ABC},$ if  $\text{AC}=\text{BC}$ and  $A{B}^2=2A{C}^2,$ then  $\mathrm{\angle }C=$ 

1. ${\mathbf{45}}^{{}^{\circ }}$ 

2. ${\mathbf{60}}^{{}^{\circ }}$ 

3. ${\mathbf{90}}^{{}^{\circ }}$ 

4. ${\mathbf{30}}^{{}^{\circ }}$ 

Ans: (c)  ${90}^{{}^{\circ }}$ 

If two sides of a triangle are congruent, then angles opposite to those sides are congruent.

So  $\mathrm{\angle }C=90{}^{\circ }$ 

10. If  $△ABC\sim \mathrm{\Delta }EDF$ and  $△ABC$ is not similar to  $△DEF,$ then which of the following is not true?

1.  $\mathbf{BC}\times \mathbf{EF}\mathbf{=}\mathbf{AC}\times \mathbf{FD}$ 

2.  $\mathbf{AB}\times \mathbf{EF}\mathbf{=}\mathbf{AC}\times \mathbf{DE}$ 

3.  $\mathbf{BC}\times \mathbf{DE}\mathbf{=}\mathbf{AB}\times \mathbf{EF}$ 

4. $\mathbf{BC}\times \mathbf{DE}\mathbf{=}\mathbf{AB}\times \mathbf{FD}$ 

Ans: c)  $BC\times DE=AB\times EF$ 

11.A certain right-angled triangle has its area numerically equal to its perimeter. The length of each side is an even integer, what is the perimeter?

1. 24 units

2. 36 units

3. 32 units

4. 30 units

Ans: 24 units

Length of each side is even integer amd also area is numerically equal its perimeter.

 $\therefore (6,8,10)$  make a Pythagorean triplet.

 $8^2+6^2={10}^2$ 

Sum of all sides =perimeter

 $6+8+10=24$ 

12. In the given figure, if  $\text{AB}||\text{CD},$ then  $x=$ 

1. 3

2. 4

3. 5

4. 6

Ans: (a) 3

From the figure it is clear that the triangles  $\mathrm{\Delta }\text{ OAB}\sim \mathrm{\Delta }\text{ OCD}$  because all the corresponding angles are equal.

So, we have,

 ${\displaystyle \frac{OA}{OC}}={\displaystyle \frac{OB}{OD}}$ 

 ${\displaystyle \frac{2x+4}{2x+1}}={\displaystyle \frac{2x+1}{3}}$ 

 $3(2x+4)=(2x+1)(2x+1)$ 

 $6x+12=4x^2+4x+1$ 

Now, by rearranging the terms to one side of the equation we get,

 $4x^2-2x-11=0$ …….(1)

Now, by using the formula of finding roots of quadratic equation

 $\text{a}{\text{x}}^{\text{2}}\text{+bx+c=0}$  we get,

 $X={\displaystyle \frac{-b\pm (\sqrt{b^2-4ac)}}{2a}}$ 

By using this formula for equation (1) we get,

$$x={\displaystyle \frac{-(-2)\pm \sqrt{{(-2)}^2+4(-11)\left(4\right)}}{2\left(4\right)}}$$

 $\text{x=}{\displaystyle \frac{\text{2 }\pm \text{ 6}\sqrt{\text{5}}}{\text{8}}}$  

Here, the two values of  $\text{x}$  are,

 $\text{x=1}\text{.92}$ 

 $\text{x=-1}\text{.42}$  

As the length cannot be negative so, the required value of  $\text{x}$  is  $1.92$ .

13.Length of an altitude of an equilateral triangle of side '  $2\text{a}$ '  $\text{cm}$ is

1. $\mathbf{3}\mathbf{\text{acm}}$ 

2. $\sqrt{\mathbf{3}}\mathbf{a}\mathbf{\text{cm}}$ 

3. ${\displaystyle \frac{\sqrt{\mathbf{3}}}{\mathbf{2}}}\mathbf{a}\mathbf{\text{cm}}$ 

4. $\mathbf{2}\sqrt{\mathbf{3}}\mathbf{a}\mathbf{\text{cm}}$ 

Ans: (b)  $\sqrt{3}a\text{cm}$ 

In the right angled  $△ABC$ 

 $A{B}^2=A{D}^2+D{B}^2$ 

 $A{D}^2=A{B}^2-D{B}^2=4a^2-a^2=3a^2$ 

 $AD=\sqrt{3}acm$ 

14. If in two triangles  $\text{ABC}$ and  $\text{PQR},{\displaystyle \frac{AB}{QR}}={\displaystyle \frac{BC}{PR}}={\displaystyle \frac{CA}{PQ}}$ 

1. $△\mathbf{PQR}\sim △\mathbf{CAB}$ 

2. $△\mathbf{PQR}\sim \mathbf{\Delta }\mathbf{ABC}$ 

3. $\mathbf{\Delta }\mathbf{CBA}\mathbf{=}△\mathbf{PQR}$ 

4. $\mathbf{\Delta }\mathbf{BCA}\sim \mathbf{\Delta }\mathbf{PQR}$ 

Ans: a)  $△PQR\sim \mathrm{\Delta }CAB$ 

15. The area of two similar triangles are  $81\text{c}{\text{m}}^2$ and  $49\text{c}{\text{m}}^2$  respectively. If the altitude of the bigger triangle is  $4.5\text{cm},$ then the corresponding altitude of the smaller triangle is

1. $\mathbf{2.5}\mathbf{\text{cm}}$ 

2. $\mathbf{2.8}\mathbf{\text{cm}}$ 

3. $\mathbf{3.5}\mathbf{\text{cm}}$ 

4. $\mathbf{3.7}\mathbf{\text{cm}}$ 

Ans: c)  $3.5\text{cm}$ 

 ${\displaystyle \frac{Areaofl\arg er△}{Areaofsmaller△}}={\left({\displaystyle \frac{altitudeofl\arg er△}{altitudeofsmaller△}}\right)}^2$ 

 ${\displaystyle \frac{81}{49}}={\displaystyle \frac{{\left(4.5\right)}^2}{x^2}}$ 

 ${\displaystyle \frac{9}{7}}={\displaystyle \frac{4.5}{x}}$ 

 $x=3.5$ 

16.In figure,  $\text{DF}||\text{BC}$ and  $\text{AD}=1\text{cm},\text{BD}=2\text{m}.$ The ratio of the area of  $△ABC$ to the area of  $△ADE$ is

1. 9:1

2. 1: 9

3. 3:1

4. none of these

Ans: (a) 9:1

${\displaystyle \frac{Areaof△ABC}{Areaof△ADF}}={\displaystyle \frac{A{B}^2}{A{D}^2}}$ 

 $AB=AB+DB=1+2=3$ 

 ${\displaystyle \frac{A{B}^2}{A{D}^2}}={\displaystyle \frac{3^2}{1}}={\displaystyle \frac{9}{1}}$ 

17.In the given figure,  $△ABC\sim △PQR,$ then the value of  $x$  and  $y$  are

1. $(\mathbf{x}\mathbf{,}\mathbf{y})\mathbf{=}(\mathbf{6}\mathbf{,}\mathbf{20})$ 

2. $\mathbf{(}\mathbf{20}\mathbf{,}\mathbf{60}\mathbf{)}$ 

3. $\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}\mathbf{(}\mathbf{3}\mathbf{,}\mathbf{10}\mathbf{)}$ 

4. none of these

Ans: (d)none of these

 ${\displaystyle \frac{AB}{PQ}}={\displaystyle \frac{AC}{PR}}={\displaystyle \frac{BC}{QR}}$ 

 ${\displaystyle \frac{24}{16}}={\displaystyle \frac{30}{x}}={\displaystyle \frac{6}{y}}$ 

 ${\displaystyle \frac{24}{16}}={\displaystyle \frac{30}{x}}\Rightarrow x=20$ 

 ${\displaystyle \frac{24}{16}}={\displaystyle \frac{6}{y}}\Rightarrow y=4$ 

 $\therefore (x,y)=(20,4)$ 

18.In figure,  $\text{P}$  and  $\text{Q}$  are points on the sides  $\text{AB}$  and AC respectively of  $△ABC$  such that  $\text{AP}=3.5\text{cm},\text{AQ}=3\text{cm}$ and $\text{QC}=6\text{cm}$ . If  $\text{PQ}=4.5\text{cm},$ then  $\text{BC}$ is

1. $\mathbf{12.5}\mathbf{\text{cm}}$ 

2. $\mathbf{5.5}\mathbf{\text{cm}}$ 

3. $\mathbf{13.5}\mathbf{\text{cm}}$ 

4. none of these

Ans: d)none of these.

 ${\displaystyle \frac{AC}{AP}}={\displaystyle \frac{CB}{PQ}}={\displaystyle \frac{AB}{AQ}}$ 

 ${\displaystyle \frac{3+6}{3.5}}={\displaystyle \frac{BC}{4.5}}\Rightarrow BC=11.5$ 

19.D, E, F are the mid-points of the sides  $\text{AB},\text{BC},$  and CA respectively of  $△ABC,$  then  ${\displaystyle \frac{\mathrm{ar}(\mathrm{\Delta }DEF)}{\mathrm{ar}(\mathrm{\Delta }ABC)}}$ is

1. 1: 4

2. 4: 1

3. 1: 2

4. none of these

Ans: (a) 1: 4

 $AB=2AD$ 

 ${\displaystyle \frac{\mathrm{ar}(\mathrm{\Delta }DEF)}{\mathrm{ar}(\mathrm{\Delta }ABC)}}={\left({\displaystyle \frac{DE}{AB}}\right)}^2$ 

 $={\left({\displaystyle \frac{1}{2}}\right)}^2={\displaystyle \frac{1}{4}}$ 

2 Marks Questions

1. In the given figures,  $△ODC\sim △OBA,\mathrm{\angle }BOC={125}^{{}^{\circ }}$  and  $\mathrm{\angle }CDO={70}^{{}^{\circ }}.$  Find

(i)  $\mathrm{\angle }\mathbf{DOC}$ 

Ans:  The sum of each interior angle and its adjacent exterior angle is equal to 180 degrees

 $\mathrm{\angle }DOC={180}^{{}^{\circ }}-{125}^{{}^{\circ }}={55}^{{}^{\circ }}$ 

(ii)  $\mathrm{\angle }\mathbf{DCO}$ 

Ans:   $\mathrm{\angle }DCO={180}^{{}^{\circ }}-({70}^{{}^{\circ }}+{55}^{{}^{\circ }})[\because DOB$ is a straight line and OC stands on it]

 $={180}^{{}^{\circ }}-{125}^{{}^{\circ }}={55}^{{}^{\circ }}[\because$ sum of angles of a triangle  $={180}^{{}^{\circ }}]$ 

(iii)  $\mathrm{\angle }\mathbf{OAB}$ 

Ans:  $\left[\begin{array}{l}\because △ODC\sim △OBA(\text{ given })\\ \therefore \mathrm{\angle }DOC=\mathrm{\angle }AOB,\mathrm{\angle }ODC=\mathrm{\angle }OBA,\mathrm{\angle }DCO=\mathrm{\angle }OAB\end{array}\right]$ 

(iv)  $\mathrm{\angle }\mathbf{AOB}$ 

Ans:  $\mathrm{\angle }AOB=\mathrm{\angle }DOC={55}^{{}^{\circ }}$ 

(v)  $\mathrm{\angle }\mathbf{OBA}$ 

Ans:  $\mathrm{\angle }OBA=\mathrm{\angle }ODC={70}^{{}^{\circ }}$

2.  $△ABC\sim △DEF$  and their areas are respectively  $64\text{c}{\text{m}}^2$  and  $121\text{c}{\text{m}}^2.$ If  $\text{EF}=15.4\text{cm},$ find  $\text{BC}.$ 

Ans: $\text{ Since }△ABC\sim \mathrm{\Delta }DEF\therefore {\displaystyle \frac{\mathrm{area}(\mathrm{\Delta }ABC)}{\mathrm{area}(\mathrm{\Delta }DEF)}}={\displaystyle \frac{B{C}^2}{E{F}^2}}$ 

 $[\because$ the ratio of the areas of two similar triangles is equal to the ratio of the squares of the corresponding sides]

 $\Rightarrow {\displaystyle \frac{64}{121}}={\displaystyle \frac{B{C}^2}{{(15.4)}^2}}$ 

 $\Rightarrow B{C}^2={\displaystyle \frac{64\times 154\times 154}{121\times 10\times 10}}={\displaystyle \frac{64\times 14\times 14}{100}}$ 

 $\Rightarrow BC={\displaystyle \frac{8\times 14}{10}}=11.2\text{cm}$ 

3.  $\text{ABC}$ is an isosceles right triangle right-angled at  $\text{B}\text{.}$ Prove that  $A{C}^2=2A{B}^2$ 

Ans: In right-angled  $△ABC,$ right  $\mathrm{\angle }AatB$ 

 $A{C}^2=A{B}^2+B{C}^2[$ By Pythagoras theorem]

 $=A{B}^2+A{B}^2=2A{B}^2[\because BC=AB(given)]$ 

 $=A{C}^2=2A{B}^2$ 

4. In the figure,  $\text{DE}||\text{AC}$  and  ${\displaystyle \frac{BE}{EC}}={\displaystyle \frac{BC}{CP}},$  prove that

Ans: In  $\mathrm{\Delta }ABC,DE\Vert AC$ 

 $\therefore {\displaystyle \frac{BD}{DA}}={\displaystyle \frac{BE}{EC}}\dots \dots (i)$  [By Thales's Theorem]

Also  ${\displaystyle \frac{BE}{EC}}={\displaystyle \frac{BC}{CP}}(given)\dots \dots .(ii)$ 

 $\therefore$ from (i) and (ii), we get

 ${\displaystyle \frac{BD}{DA}}={\displaystyle \frac{BC}{CP}}\therefore DC\Vert AP$ 

 (By the converse of Thales's Theorem)

5.The hypotenuse of a right triangle is  $6\text{m}$  more than twice of the shortest side. If the third side is  $2\text{m}$  less than the hypotenuse. Find the side of the triangle.

Ans: Let shortest side be  $Xm$  in length

Then hypotenuse  $=(2x+6)m$ 

And third side  $=(2x+4)m$ 

We have,

 ${(2x+6)}^2=x^2+{(2x+4)}^2$ 

 $\Rightarrow 4x^2+24x+36=x^2+4x^2+16+16x$ 

 $\Rightarrow x^2-8x-20=0$ 

 $\Rightarrow x=10\text{ or }x=-2$ 

 $\Rightarrow x=10$ 

Hence, the sides of triangle are  $10\text{m},26\text{m}$  and  $24\text{m}.$ 

6. PQR is a right triangle right angled at P and M is a point on QR such that $$\text{PM}\mathrm{\perp }$$QR. Show that $$P{M}^2=QM.MR$$

Ans: Given PQR is a right triangle right angled at P and  $PM\mathrm{\perp }QR$ 

 $\therefore △PMR\sim △PMQ$  

 $\therefore {\displaystyle \frac{PR}{PQ}}={\displaystyle \frac{PM}{QM}}={\displaystyle \frac{MR}{PM}}$ 

 $\Rightarrow {\displaystyle \frac{PM}{QM}}={\displaystyle \frac{MR}{PM}}$ 

 $\text{ i}\text{. e}\text{., }P{M}^2=QM.MR$ 

7. In the given figure, $$\mathbf{D}\mathbf{E}\Vert \mathbf{O}\mathbf{Q}\mathbf{and}\mathbf{D}\mathbf{F}\Vert OR,$$Prove that $$\mathbf{E}\mathbf{F}\Vert OQ$$

Ans: In  $△OQR,DE\Vert OQ$ 

 ${\displaystyle \frac{PE}{EQ}}={\displaystyle \frac{PD}{DO}}.....\left(1\right)$ 

In  $△OPR,DF\Vert OR$ 

 ${\displaystyle \frac{PD}{DO}}={\displaystyle \frac{PF}{FR}}\dots \left(ii\right)$ 

From (i) and (ii), we get

 ${\displaystyle \frac{PE}{EQ}}={\displaystyle \frac{PF}{FR}}$ 

 $\therefore From△PQR$ 

 $EF\Vert QR$ 

8. In figure,  $\text{DE}||\text{BC}$ . Find EC

Ans:  $\because DE\Vert BC$ 

 $\therefore {\displaystyle \frac{AD}{DB}}={\displaystyle \frac{AE}{EC}}$ 

 $\Rightarrow {\displaystyle \frac{1.5}{3}}={\displaystyle \frac{1}{EC}}$ 

 $\therefore EC=2\text{cm}$ 

9. In the given figure,  $\text{ABC}$ and AMP are two right-angled triangles, right angled at  $\text{B}$ and M respectively, prove that

 $(i)△ABC\sim △AMP$ 

 (ii)  ${\displaystyle \frac{CA}{PA}}={\displaystyle \frac{BC}{MP}}$ 

Ans: In  $△ABC$ and DAMP,

 $\mathrm{\angle }B=\mathrm{\angle }M($ 

Each  ${90}^{{}^{\circ }})$   

 $\mathrm{\angle }A=\mathrm{\angle }A(\text{common})$  

 $\therefore \mathrm{\angle }ACB=\mathrm{\angle }APM$ 

 $\therefore \mathrm{\Delta }S$ are equiangular 

i.e.,  $△ABC\sim △AMP$ 

 $\therefore {\displaystyle \frac{BC}{MP}}={\displaystyle \frac{CA}{PA}}$ 

10. In the given figure,  $\text{OA}\times \text{OB}=\text{OC}\times OD$  or  ${\displaystyle \frac{OA}{OC}}={\displaystyle \frac{OD}{OB}}$ , prove that  $\mathrm{\angle }A=\mathrm{\angle }Cand\mathrm{\angle }B=\mathrm{\angle }D$ 

Ans: In  $△AOD$ and  $△BOC,$ 

 $OA\times OB=OC\times OD$ 

 $\text{ i}\text{.e }{\displaystyle \frac{OA}{OC}}={\displaystyle \frac{OD}{OB}}$ 

And  $\mathrm{\angle }AOD=\mathrm{\angle }BOC$  (Vertically opposite Angles)

 $\therefore △AOD\sim \mathrm{\Delta }BOC(BySAS)$ 

 $\therefore \mathrm{\angle }A=\mathrm{\angle }C$  and  $\mathrm{\angle }B=\mathrm{\angle }D$  (Corresponding angles of similar  $\mathrm{\Delta }$ )

11. In the given figure, 

 $\text{DE}\Vert \text{BC}$ and $\text{AD}=1\text{cm},\text{BD}=2\text{cm}.$ What is the ratio of the area of  $△ABC$ to the area of $△ADE$ ?

Ans:  $\because DE\Vert BC$ in  $△ABC$  $\therefore \mathrm{\angle }ADE=\mathrm{\angle }ABC$ 

 $\mathrm{\angle }AED=\mathrm{\angle }ACB$ 

Also  $\mathrm{\angle }DAE=\mathrm{\angle }DAC$ 

 $\therefore △ADE\sim \mathrm{\Delta }ABC$ 

 $\therefore △ADE\sim △ABC$ 

 $\therefore {\displaystyle \frac{A{D}^2}{A{B}^2}}={\displaystyle \frac{\mathrm{area}(△ADE)}{\mathrm{area}(△ABC)}}$ 

 $\Rightarrow {\displaystyle \frac{1^2}{3^2}}={\displaystyle \frac{\mathrm{area}(\mathrm{\Delta }ADE)}{\text{ area }(\mathrm{\Delta }ABC)}}[\because AB=AD+OB=1+2=3]$ 

 $\text{ Hence, }{\displaystyle \frac{\mathrm{area}(\mathrm{\Delta }ABC)}{\mathrm{area}(\mathrm{\Delta }ADE)}}={\displaystyle \frac{9}{1}}$ 

12. A right-angle triangle has hypotenuse of length  $\text{p }cm$  and one side of length  $\text{qcm}.$ If  $\mathbf{p}-q=1,$ Find the length of third side of the triangle.

Ans: Let third side  $=\text{xcm}$ 

Then by Pythagoras theorem.

 $p^2=q^2+x^2$ 

 $x^2=p^2-q^2$  

 $=(p+q)\times 1(\because p-q=1)$ 

$$=q+1+q$$

$$=2q+1$$

 $\therefore x=\sqrt{2q+1}$ 

13. The length of the diagonals of a rhombus are  $24\text{cm}$ and $10\text{cm}$ . Find each side of rhombus.

Ans:  $AC=24\text{cm}\therefore AO=12\text{cm}$ 

 $BD=10\text{cm}\therefore OD=5\text{cm}$ 

From right-angled  $△AODA{D}^2=A{O}^2+O{D}^2$ 

 $\Rightarrow A{D}^2={12}^2+5^2$ 

 $\Rightarrow A{D}^2=169$ 

 $\Rightarrow AD=13\text{cm}$ 

Hence each side  $=13\text{cm}$ 

14. In an isosceles right-angled triangle, prove that hypotenuse is  $\sqrt{2}$ times 

Ans: Let hypotenuse of right-angled  $\mathrm{\Delta }=h$ units and equal sides of triangle  $x$ units

 $\therefore$ By Pythagoras theorem,

 $h^2=x^2+x^2$ 

 $\Rightarrow h^2=2x^2$ 

 $\Rightarrow h=\sqrt{2}x$ 

15. In figure, express  $\text{x}$  in terms of  $\mathbf{a},\mathbf{b},\mathbf{c}.$ 

Ans:  $△ABO\sim △OCD$ 

 $\Rightarrow {\displaystyle \frac{x}{a}}={\displaystyle \frac{x+b}{c}}$ 

 $\Rightarrow x=ax+ab$ 

 $\Rightarrow x(c-a)=ab$ 

 $\Rightarrow x={\displaystyle \frac{ab}{c-a}}$ 

16. The perimeter of two similar triangle ABC and PQR are respectively  $36\text{cm}$ and  $24\text{cm}.$ If  $\text{PQ}=10\text{cm},$  find  $\text{AB}.$ 

Ans:  $△ABC\sim \mathrm{\Delta }PQR$ 

 $\therefore {\displaystyle \frac{AB}{PQ}}={\displaystyle \frac{BC}{QR}}={\displaystyle \frac{AC}{PR}}$ 

 $\Rightarrow {\displaystyle \frac{AB+BC+AC}{PQ+QR+PR}}={\displaystyle \frac{\text{ perimeter of }\mathrm{\Delta }ABC}{\text{ perimeter of }\mathrm{\Delta }PQR}}$ 

 $\Rightarrow {\displaystyle \frac{AB}{10}}={\displaystyle \frac{36}{24}}$ 

 $\Rightarrow AB={\displaystyle \frac{36\times 10}{24}}=15\text{cm}$ 

17. In the given figure,  $\text{DE}||\text{BC}.$  If  $AD=x,DB=x-2,AE=x+2,EC=x-1$ find the value of $$\mathbf{x}.$$

Ans: In the given figure,

$$DE\Vert BC$$

 $\therefore {\displaystyle \frac{AD}{DB}}={\displaystyle \frac{AE}{EC}}$ 

 $\Rightarrow {\displaystyle \frac{x}{x-2}}={\displaystyle \frac{x+2}{x-1}}$ 

 $\Rightarrow x^2-x=x^2-4$ 

 $\Rightarrow x=4$ 

18. In the given figure,  ${\displaystyle \frac{AO}{OC}}={\displaystyle \frac{BO}{OD}}={\displaystyle \frac{1}{2}}$ and  $\text{AB}=5\text{cm},$ find the value of  $\text{DC}.$ 

Ans: In  $△AOB$ and  $△COD,$ 

 $\mathrm{\angle }AOB=\mathrm{\angle }COD$  [Vertically opposite angles]

 ${\displaystyle \frac{AO}{OC}}={\displaystyle \frac{BO}{OD}}\Rightarrow {\displaystyle \frac{AO}{OB}}={\displaystyle \frac{OC}{OD}}$  [Given]

 $\therefore △AOB\sim △COD$  [By SAS similarity]

 $\therefore {\displaystyle \frac{AO}{CO}}={\displaystyle \frac{BO}{DO}}={\displaystyle \frac{AB}{CD}}$ 

 ${\displaystyle \frac{1}{2}}={\displaystyle \frac{AB}{DC}}[{\displaystyle \frac{AO}{OC}}={\displaystyle \frac{BO}{OD}}={\displaystyle \frac{1}{2}}isgiven]$ 

 $\Rightarrow {\displaystyle \frac{1}{2}}={\displaystyle \frac{5}{DC}}$ 

 $\Rightarrow DC=10\text{cm}$ 

19. In  $△ABC,AB=AC$ and  $\text{D}$ is a point on side  $\text{AC},$ such that  $\text{B}{\text{C}}^2=AC\times CD\cdot$ Prove that  $\text{BD}=\mathbf{BC}.$ 

Ans: Given:  $\mathrm{\Delta }ABC$ in which  $AB=AC,D$ is a point on  $BC$ 

To prove:  $\text{BD}=\text{BC }$ 

Proof:  $B{C}^2=AC\times CD$ [given] 

 $\Rightarrow {\displaystyle \frac{BC}{AC}}={\displaystyle \frac{DC}{BC}}$ 

 $\text{ In }△ABC\text{ and }△BDC,$ 

 $\Rightarrow {\displaystyle \frac{BC}{CA}}={\displaystyle \frac{DC}{CB}}\text{ and }\mathrm{\angle }C=\mathrm{\angle }C[\text{ Common }]$ 

 $\therefore △ABC\sim △BDC[\text{SASsimilarity}]$ 

 $\Rightarrow {\displaystyle \frac{AB}{BD}}={\displaystyle \frac{AC}{BC}}\Rightarrow {\displaystyle \frac{AC}{BD}}={\displaystyle \frac{AC}{BC}}[\because AB=AC]$ 

 $\Rightarrow BD=BC$ 

3 Marks Questions

1. In the given figure,  ${\displaystyle \frac{QT}{PR}}={\displaystyle \frac{QR}{QS}}$  and  $\mathrm{\angle }1=\mathrm{\angle }2$ . Prove that  $\mathrm{\Delta }PQS\sim \mathrm{\Delta }TQR$ 

Ans: Since  ${\displaystyle \frac{QT}{PR}}={\displaystyle \frac{QR}{QS}}$ [Given]

 $\therefore {\displaystyle \frac{QT}{QR}}={\displaystyle \frac{PR}{QS}}$ 

Since  $\mathrm{\angle }1=\mathrm{\angle }2[Given]$ 

 $PQ=PR$ 

 (In  $\mathrm{\Delta }PQR$ sides opposites to opposite angles are equal)

 $\therefore {\displaystyle \frac{QT}{QR}}={\displaystyle \frac{PQ}{QS}}\dots \dots .(iii)[\mathrm{Form}(i)\text{ and }(ii)]$ 

Now in  $△PQS$ and  $△TQR$ 

From (iii), $${\displaystyle \frac{PQ}{QS}}={\displaystyle \frac{QT}{QR}}i.e.{\displaystyle \frac{PQ}{QT}}={\displaystyle \frac{QS}{QR}}$$

And  $\mathrm{\angle }Q=\mathrm{\angle }Q$ [Common]

 $\therefore \mathrm{\Delta }PQS\sim \mathrm{\Delta }TQR$  [By S.A.S. Rule of similarity]

2. In the given figure, PA, QB and RC are each perpendicular to AC. Prove that  ${\displaystyle \frac{1}{x}}+{\displaystyle \frac{1}{2}}={\displaystyle \frac{1}{y}}$ 

Ans: In   $△PAC\text{ and }\mathrm{\Delta }QBC$ 

 $\mathrm{\angle }PAC=\mathrm{\angle }QBC[\text{ Each }={90}^{{}^{\circ }}]$ 

 $\mathrm{\angle }PCA=\mathrm{\angle }QCB[\text{Common}]$ 

 $\therefore △PAC\sim \mathrm{\Delta }QBC$ 

 ${\displaystyle \frac{x}{y}}={\displaystyle \frac{AC}{BC}}\text{ i}\text{.e}\text{. }{\displaystyle \frac{y}{x}}={\displaystyle \frac{BC}{AC}}\dots \dots .(i)$ 

 $\text{ Similarly, }{\displaystyle \frac{z}{y}}={\displaystyle \frac{AC}{AB}}\text{ i}\text{.e}\text{. }{\displaystyle \frac{y}{z}}={\displaystyle \frac{AB}{AC}}\dots \dots .(ii)$ 

 $\text{Adding (i) and (ii), we get }$ 

 $\Rightarrow {\displaystyle \frac{BC+AB}{AC}}={\displaystyle \frac{y}{x}}+{\displaystyle \frac{y}{z}}=y({\displaystyle \frac{1}{x}}+{\displaystyle \frac{1}{z}})$ 

 $\Rightarrow {\displaystyle \frac{AC}{AC}}=y({\displaystyle \frac{1}{x}}+{\displaystyle \frac{1}{z}})\Rightarrow 1=({\displaystyle \frac{1}{x}}+{\displaystyle \frac{1}{z}})$ 

 $\Rightarrow {\displaystyle \frac{1}{y}}={\displaystyle \frac{1}{x}}+{\displaystyle \frac{1}{z}}$ 

3. In the given figure,  $\text{DE}||\text{BC}$ and AD:DB $$=\mathbf{5}:\mathbf{4},$$find  ${\displaystyle \frac{\text{ area }(\mathrm{\Delta }DFE)}{\text{ area }(\mathrm{\Delta }CFB)}}$ .

Ans: In  $△ADE$ and  $△ABC,$   $\mathrm{\angle }1=\mathrm{\angle }1$ (Common)

 $\mathrm{\angle }2=\mathrm{\angle }ACB$  [Corresponding angles]

 $\therefore △ADE\sim △ABC$  [By A.A Rule]

 $\therefore {\displaystyle \frac{DE}{BC}}={\displaystyle \frac{AD}{AB}}\dots \dots ..i)$ 

Again in  $△DEF$ and  $△CFB,$ 

 $\mathrm{\angle }3=\mathrm{\angle }6[Alternateangles]$ 

 $\mathrm{\angle }4=\mathrm{\angle }5$  [Vertically oppositeangles]

 $\therefore \mathrm{\Delta }DFE\sim △CFB$  [By A.A Rule]

 $\therefore {\displaystyle \frac{\text{ Area }(\mathrm{\Delta }DFE)}{\text{ area }(\mathrm{\Delta }CFB)}}={\displaystyle \frac{D{E}^2}{B{C}^2}}={\left({\displaystyle \frac{AD}{AB}}\right)}^2$  [From (i)]

 $={\left({\displaystyle \frac{5}{9}}\right)}^2[\because {\displaystyle \frac{AD}{DB}}={\displaystyle \frac{5}{4}}\Rightarrow {\displaystyle \frac{AD}{AD+DB}}={\displaystyle \frac{5}{5+4}}\Rightarrow {\displaystyle \frac{AD}{DB}}={\displaystyle \frac{5}{9}}]$ 

 $\therefore {\displaystyle \frac{\mathrm{area}(\mathrm{\Delta }DFE)}{\mathrm{area}(\mathrm{\Delta }CFB)}}={\displaystyle \frac{25}{81}}$ 

3. Determine the length of an altitude of an equilateral triangle of side ‘2a’ cm.

Ans: In right triangles  $△ADB$ and  $△ADC,$ 

$$AB=AC$$ 

$$AD=AD$$

 $\therefore \mathrm{\angle }ADB=\mathrm{\angle }ADC(Each={90}^{{}^{\circ }})$ 

 $\therefore △ADB\cong △ADC(RHS)$ 

 $\therefore BD=DC(CPCT)$ 

 $\therefore BD=DC=a[\because BC=2a]$ 

In right  $△ADB,A{D}^2+B{D}^2=A{B}^2$ (By Pythagoras Theorem)

 $\Rightarrow A{D}^2+a^2={(2a)}^2$ 

 $\Rightarrow A{D}^2=4a^2-a^2=3a^2$