Chapter 10 Areas of Parallelograms and Triangles Ex 10.2
Question 1.
In figure, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.
Solution:
We know that,
Area of parallelogram = Base x Altitude
Given, AE = 8 cm CF = 10 cm and AB = 16cm
∴ ar (parallelogramABCD) = DC x AE
= 16 x 8 cm2 (∵ AE = 8 cm)…(i)
and ar (parallelogram ABCD) = AD x CF – AD x 10 ( ∵ CF = 10 cm)
From Eq. (i), we have,
16 x 8 = AD x 10
Question 2.
Solution:
Given: E,F, G and H are respectively the mid-points of the sides AB, BC, CD and AD. Joint if, it will parallel to CD and AB.
Now, parallelogram HDCF and triangle HGF stand on the same base HF and lie between the same parallel lines DC and HF.
Question 3.
P and Q are any two points lying on the sides DC and AD, respectively of a parallelogram ABCD. Show that ar (APB) = ar(BQC).
Solution:
Given: a parallelogram ABCD. P and Q are any two points lying on the sides DC and AD, respectively.
Question 4.
In figure, P is a point in the interior of a parallelogram ABCD. Show that
(ii) ar (APD) + ar(PBC) = ar (APB) + ar (PCD)
Solution:
Given: ABCD is a parallelogram. So, AB || CD, AD |[ BC.
(i) Now, draw MPR parallel to AB and CD both and also draw a perpendicular PS on AB.
∵ MR || AB and AM || BR
Question 5.
In figure, PQRS and ABRS are parallelograms and X is any point on side BR. Show that
(i) ar (PQRS) = ar (ABRS)
Solution:
Question 6.
A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the fields is divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should she do it.
Solution:
Given: PQRS is a parallelogram and A is any point as RS. Now, join PA and PQ. Thus, the field will be divided into three parts and each part is in the shape of a triangle.
Since, the AAPQ and parallelogram PQRS lie on the same base PQ and between same parallel lines PQ and SR.
