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Published on: 16/09/2019
Areas of Parallelograms and Triangles
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1.
In given figure, ABCD is a parallelogram and BE\(\bot \)AD. If BE=14 cm and AD=8 cm, find the area of \(\Delta \)DBC.
2.
In the figure, PS||QR, Show that ar(\(\Delta \)ROS) = ar(\(\Delta \)POQ).
3.
ABCD is a parallelogram with area 80 sq. cm. The diagonals AC and BD intersect at O. P is the mid-point of OA. Calculator ar (\(\Delta\)BOP).
4.
ABCD is a quadrilateral and BD is one of its diagonals as shown in figure. Show that ABCD is a parallelogram and find its area.
5.
In the figure, PQRS is parallelogram with PQ=8 cm and ar(\(\Delta \)PXQ)=32 cm2. Find the altitude of PQRS and hence its area.
6.
In the given figure, ABCD is a \(BE\bot AD\)parallelogram. If BE = 14 cm and AD = 8 cm, find the area of \(\Delta\)DBC.
7.
ABCD is a parallelogram with area 80 sq. cm. The diagonals AC and BD intersect at O. P is the midpoint of OA. Calculate ar(\(\Delta\) BOP)
8.
In the given figure:
(a) name the two triangles on the same base AB and between the same parallels.
(b) name a triangle and a parallelogram on the same base BC and between the same parallels.
9.
In the figure, ABCD is a parallelogram. P is a point on AB produced and \(DN\bot AB\) . If AB = 8 cm and DN = 3 cm. Find the area of \(\Delta \) CPD.
10.
If the area of parallelogram (shown in figure) is 80 cm2 , then find area of \(\Delta \) ADP.
11.
In the figure, ABCD is a parallelogram. AB = 12 cm, DM = 6 cm and BN = 9 cm. Find the length of AD.
12.
In the figure, ar( \(\Delta\)ABE) = 50 cm2. Find the area of the parallelogram ABCD. Give reasons.
13.
In the given figure, ABC and DBC are triangles on the same base and between parallel lines I and m. If AB = 3 cm, BC = 5 cm, \(\angle A=90°\), find area of \(\Delta \) DBC.
14.
In a parallelogram ABCD, AB = 8 cm. The altitudes corresponding to sides AB and AD are respectively 4 cm and 5 cm. Find measure of AD.
15.
In the figure, PQRS is a parallelogram with PQ = 12 cm, altitudes corresponding to PQ and SP are respectively 8 cm and 10 cm. Find SP.
1.
Given: BE = 14 cm, AD = 8 cm
\(\therefore \ Area(\triangle ADB)=\frac { 1 }{ 2 } \times 8\times 14\)
= 56 cm2
\(\because\) ABCD is a parallelogram.
\(\therefore\) ar(\(\Delta \)DBC) = ar(\(\Delta \)ADB) = 56 cm2.
2.
\(\Delta \)PSR and \(\Delta \)PSQ are on the same base PS and between the same parallels PS and QR.
ar(\(\Delta \)PSR) = ar(\(\Delta \)PSQ)
ar(\(\Delta \)PSR) - ar(\(\Delta \)PSO) = ar(\(\Delta \)PSQ) - ar(\(\Delta \)PSO)
ar(\(\Delta \)ROS) = ar(\(\Delta \)POQ).
3.
Area of a parallelogram is divided into four equal parts by the diagonals.
\(\Rightarrow \ ar(AOB)=\frac { 1 }{ 4 } ar(ABCD)\)
\(ar(\Delta BOP)=\frac { 1 }{ 2 } ar(AOB)=\frac { 1 }{ 8 } ar\ (ABCD)\)
\(=\frac { 1 }{ 8 } \times 80=10\quad { cm }^{ 2 }\)
4.
In the figure,
\(\angle CDB=\angle ABD=90°\)
But they are alternate angles.
\(\therefore\) AB || DC
Also DC = AB = 3 cm.
A quadrilateral with a pair of equal and parallel sides is a parallelogram.
\(\therefore\) Area = b\(\times\)h
=(3\(\times\)4) cm2
= 12 cm2
5.
Area of \((\Delta PXO)=\frac { 1 }{ 2 } \times b\times h\)
(b=base, h=height)
\(\Rightarrow \ 32=\frac { 1 }{ 2 } \times PQ\times h\)
\(\Rightarrow \ 32=\frac { 1 }{ 2 } \times 8\times h\)
\(\Rightarrow \ h=\frac { 32\times 2 }{ 8 } \)
\(\Rightarrow\) h = 8 cm
now, area of ||gm PQRS = b\(\times\)h
= 8 \(\times\) 8
= 64 cm2
Hence altitude of PQRS = 8 cm
and area of PQRS = 64 cm2
6.
56 cm2
7.
10 cm2
8.
\((a)\Delta DAB,DCAB\)
\(\\ (b)\Delta DAB,||gm\ ABCD\)
9.
12 cm2
10.
40 cm2
11.
8 cm
12.
100 cm2
13.
7.5 cm2
14.
6.4cm
15.
9.6 cm
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