Question

In \[\vartriangle ABC\], P, Q, R are the mid-points of sides AB, BC and AC respectively.
If \[ABC = 120\]\[c{m^2}\], find PQR, PQCR and PBCR.

Answer 1

Hint: We draw a triangle and join the midpoints of each side to each other forming 4 triangles within the triangle. Using the mid-point theorem we write the value of parallel lines joining the mid-points in terms of the side of the triangle. Use SAS congruence rule to prove all four triangles congruent to each other. From the property that congruent triangles have equal area we find the area of a single triangle and use it to find the required areas.

Complete step-by-step answer:
We draw a triangle ABC where P, Q and R are the mid-points of the sides AB, BC and CA respectively.

Since, P is the midpoint of AB and then P must divide the side AB into two equal halves.
\[ \Rightarrow AP = BP\]
Also, Q is the midpoint of BC and then Q must divide the side BC into two equal halves.
\[ \Rightarrow BQ = CQ\]
And R is mid-point of AC and then R must divide the side AC into two equal halves.
\[ \Rightarrow AR = CR\]
Now we know from the mid-point theorem that the line joining midpoints of any two sides is parallel to the third side of the triangle and is half the length of the third side of the triangle.
We know PQ is line joining midpoints of sides AB and BC then,\[PQ\parallel AC\]
\[ \Rightarrow PQ = \dfrac{1}{2}AC\]
We know QR is line joining midpoints of sides BC and AC then,\[QR\parallel AB\]
\[ \Rightarrow QR = \dfrac{1}{2}AB\]
We know PR is line joining midpoints of sides AB and AC then,\[PR\parallel BC\]
\[ \Rightarrow PR = \dfrac{1}{2}BC\]
We have four triangles\[\vartriangle APR,\vartriangle PBQ,\vartriangle RQC,\vartriangle PQR\]. We will show triangles congruent to each other.
First we take \[\vartriangle ARP\]and\[\vartriangle RCQ\]
\[PR = QC\](As both lengths are equal to half of length of BC)
\[AR = RC\](As R is the midpoint of AC)
\[\angle ARP = \angle RCQ\](As corresponding angles are equal when parallel lines PR and BC are cut by transversal AC)
Then by SAS congruence rule \[\vartriangle ARP \cong \vartriangle RCQ\]
Then by property of congruent triangles
Area of \[\vartriangle ARP\]\[ = \] Area of \[\vartriangle RCQ\] … (1)
Now we take \[\vartriangle APR\]and\[\vartriangle PBQ\]
\[PR = BQ\](As both lengths are equal to half of length of BC)
\[AP = PB\](As P is the mid-point of AB)
\[\angle APR = \angle PBQ\](As corresponding angles are equal when parallel lines PR and BC are cut by transversal AC)
Then by SAS congruence rule \[\vartriangle APR \cong \vartriangle PBQ\]
Then by property of congruent triangles
Area of \[\vartriangle APR\]\[ = \] Area of \[\vartriangle PBQ\] … (2)
Now we take \[\vartriangle PRQ\]and\[\vartriangle CQR\]
\[PR = QC\](As both lengths are equal to half of length of BC)
\[PQ = RC\](As both lengths are equal to half of length of AC)
\[QR = QR\]
Then by SSS congruence rule \[\vartriangle PRQ \cong \vartriangle CQR\]
Then by property of congruent triangles
Area of \[\vartriangle PRQ\]\[ = \] Area of \[\vartriangle CQR\] … (3)
From equations (1), (2) and (3)
Area \[\vartriangle APR\]\[ = \]Area \[\vartriangle PBQ\]\[ = \]Area \[\vartriangle PQR\]\[ = \] Area \[\vartriangle CQR\] … (4)
We know the triangle ABC is formed by 4 triangles APR, PBQ, PQR and CQR
\[ \Rightarrow \]Area\[\vartriangle ABC\]\[ = \]Area\[\vartriangle APR\]\[ + \]Area\[\vartriangle PBQ\]\[ + \]Area\[\vartriangle PQR\]\[ + \]Area\[\vartriangle CQR\]
Substitute all areas in RHS of the equation to Area\[\vartriangle PQR\]
\[ \Rightarrow \]Area\[\vartriangle ABC\]\[ = \]Area\[\vartriangle PQR\]\[ + \]Area\[\vartriangle PQR\]\[ + \]Area\[\vartriangle PQR\]\[ + \]Area\[\vartriangle PQR\]
\[ \Rightarrow \]Area\[\vartriangle ABC\]\[ = 4 \times \]Area\[\vartriangle PQR\]
Substitute the value of Area\[\vartriangle ABC\]as \[120\]in LHS of the equation
\[ \Rightarrow 120 = 4 \times \] Area\[\vartriangle PQR\]
Divide both sides by 4
\[ \Rightarrow \dfrac{{120}}{4} = \dfrac{4}{4} \times \] Area\[\vartriangle PQR\]
Cancel same factors from numerator and denominator
\[ \Rightarrow 30 = \] Area\[\vartriangle PQR\]
So, area of\[\vartriangle PQR\]is 30\[c{m^2}\]
 Area \[\vartriangle APR\]\[ = \]Area \[\vartriangle PBQ\]\[ = \]Area \[\vartriangle PQR\]\[ = \] Area \[\vartriangle CQR\]\[ = \]30\[c{m^2}\]
Now Area PQCR\[ = \]Area \[\vartriangle PQR\]\[ + \] Area \[\vartriangle CQR\]
\[ \Rightarrow \]Area PQCR\[ = 2 \times \]Area \[\vartriangle PQR\](From equation (4))
Substitute the value of Area\[\vartriangle PQR\]as 30
\[ \Rightarrow \]Area PQCR\[ = 2 \times 30\]\[c{m^2}\]
\[ \Rightarrow \]Area PQCR\[ = 60\]\[c{m^2}\]
Now Area PBCR\[ = \]Area \[\vartriangle PQR\]\[ + \] Area \[\vartriangle CQR\]\[ + \]Area\[\vartriangle PBQ\](from equation (4))
\[ \Rightarrow \]Area PQCR\[ = 3 \times \]Area \[\vartriangle PQR\]
Substitute the value of Area\[\vartriangle PQR\]as 30
\[ \Rightarrow \]Area PQCR\[ = 3 \times 30\]\[c{m^2}\]
\[ \Rightarrow \]Area PQCR\[ = 90\]\[c{m^2}\]

Thus, PQR is 30\[c{m^2}\], PQCR is 60\[c{m^2}\] and PBCR is 90\[c{m^2}\].

Note: Students might get confused in proving SAS congruence as they write any angle of one triangle equal to corresponding angle of another triangle. Keep in mind the including angle between the equal sides should be taken i.e. the angle between the sides.
* Mid-point theorem: Line joining midpoints of two sides of a triangle is parallel to the third side of the triangle and is half the length of the third side of the triangle.
* SAS congruence rule states that two triangles are congruent is two sides and the included angle of a triangle is equal to corresponding two sides and the corresponding angle of the other triangle.