Areas of Parallelograms and Triangles CBSE Class 9 Mathematics Worksheets Set 10

Practice Set

Question 1. \( ABCD \) is a parallelogram; \( E,F \) are the mid-points of \( AB \) and \( CD \). A line \( GH \) meets \( AD \) at \( G \), \( EF \) at \( P \) and \( BC \) at \( H \). Prove \( GP=PH \).
Answer:
\( AE=EB \) and \( AD\parallel EF\parallel BC \), so the three parallel lines make equal intercepts on \( AB \).
By the equal intercepts theorem they make equal intercepts on every transversal, so \( GP=PH \).

Question 2. In parallelogram \( ABCD \) a line through \( A \) meets \( DC \) at \( P \) and \( BC \) produced at \( Q \). Prove \( \text{ar}(\triangle BPC)=\text{ar}(\triangle DPQ) \).
Answer:
\( AD\parallel CQ \), so \( \triangle ADQ \) and \( \triangle ADC \) (same base \( AD \)) have equal area.
Subtract \( \text{ar}(\triangle ADP) \): \( \text{ar}(\triangle DPQ)=\text{ar}(\triangle APC) \).
\( AB\parallel PC \), so \( \text{ar}(\triangle APC)=\text{ar}(\triangle BPC) \) (same base \( PC \)). Hence \( \text{ar}(BPC)=\text{ar}(DPQ) \).

Question 3. \( P \) and \( Q \) are points on sides \( DC \) and \( AD \) of parallelogram \( ABCD \). Show \( \text{ar}(\triangle APB)=\text{ar}(\triangle BQC) \).
Answer:
\( \triangle APB \) and the parallelogram have the same base \( AB \) and lie between \( AB\parallel DC \), so \( \text{ar}(APB)=\frac12\text{ar}(ABCD) \).
\( \triangle BQC \) and the parallelogram have the same base \( BC \) and lie between \( BC\parallel AD \), so \( \text{ar}(BQC)=\frac12\text{ar}(ABCD) \). Hence they are equal.

Question 4. In \( \triangle ABC \), \( AD \) is a median and \( E \) is the mid-point of \( AD \). \( BE \) produced meets \( AC \) at \( F \). Prove \( AF=\frac13AC \).
Answer:
Draw \( DM\parallel BF \) meeting \( AC \) at \( M \).
In \( \triangle BCF \), \( D \) is the mid-point of \( BC \) and \( DM\parallel BF \), so \( M \) is the mid-point of \( CF \): \( FM=MC \).
In \( \triangle ADM \), \( E \) is the mid-point of \( AD \) and \( EF\parallel DM \), so \( F \) is the mid-point of \( AM \): \( AF=FM \).
So \( AF=FM=MC \) and \( AC=3AF \).

Question 5. \( ABC \) is a triangle right-angled at \( B \) and \( P \) is the mid-point of \( AC \). Prove \( PB=PA=\frac12AC \).
Answer:
Through \( P \) draw \( PD\parallel BC \) meeting \( AB \) at \( D \). Then \( D \) is the mid-point of \( AB \) and \( PD\perp AB \).
\( \triangle PDA\cong\triangle PDB \) (SAS: \( DA=DB \), right angles, \( PD \) common), so \( PA=PB \).
Since \( PA=\frac12AC \), \( PB=PA=\frac12AC \).

Question 6. The diagonals of a parallelogram \( ABCD \) meet at \( O \). A line through \( O \) meets \( AD \) at \( P \) and \( BC \) at \( Q \). Show that \( PQ \) divides the parallelogram into two parts of equal area.
Answer:
\( OA=OC \), \( \angle AOP=\angle COQ \) (vertically opposite) and \( \angle OAP=\angle OCQ \) (alternate angles), so \( \triangle AOP\cong\triangle COQ \).
The part \( ABQP \) equals \( \triangle ABC \) minus \( \triangle COQ \) plus \( \triangle AOP \), so it has area \( \frac12\text{ar}(ABCD) \). The other part has the same area.

Question 7. \( ABCD \) is a quadrilateral. A line through \( D \) parallel to \( AC \) meets \( BC \) produced at \( P \). Prove that \( \text{ar}(\triangle ABP)=\text{ar}(ABCD) \).
Answer:
\( \triangle ACD \) and \( \triangle ACP \) are on the same base \( AC \) and between the parallels \( AC \) and \( DP \): \( \text{ar}(ACD)=\text{ar}(ACP) \).
Add \( \text{ar}(\triangle ABC) \) to both: \( \text{ar}(ABCD)=\text{ar}(ABP) \).

Question 8. The area of a triangle equals the area of a rectangle 18 cm by 12 cm. The base of the triangle is 24 cm. Find its altitude.
Answer:
Area of the rectangle \( =18\times12=216 \) cm\( ^2 \).
\( \frac12\times24\times h=216 \), so \( h=18 \) cm.