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ML Aggarwal Solutions Class 9 Mathematics Solutions for Theorems on Area Exercise 14 in Chapter 14 - Theorems on Area
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(a) In the figure (1) given below, the area of parallelogram ABCD is 29 \mathrm{cm}^{2}
Calculate the height of parallelogram ABEF if AB = 5.8 cm
(b) In figure (2) given below, the area of ∆ABD is 24 sq. units. If AB = 8 units, find
the height of ABC.
(c) In figure (3) given below, E and F are midpoints of sides AB and CD
respectively of parallelogram ABCD. If the area of parallelogram ABC is 36 \mathrm{cm}^{2}
(i) State the area of ∆ APD.
(ii) Name the parallelogram whose area is equal to the area of ∆ APD.
(a) Given:
From fig (1)
ar ||gm ABCD = 29 \mathrm{cm}^{2}
To find:
Height of parallelogram ABEF if AB = 5.8 cm
Now, let us find
We know that ||gm ABCD and ||gm ABEF with equal bases and between the same
parallels so that their area is the same.
ar (||gm ABEF) = ar (||gm ABCD)
ar (||gm ABEF) = 29 \mathrm{cm}^{2} 2 ….. (1) [Since, ar ||gm ABCD = 29\mathrm{cm}^{2}
also, ar (||gm ABEF = base × height)
29 = AB × height [From (1)]
29 = 5.8 × height
Height = 29/5.8
= 5
∴ The height of parallelogram ABEF is 5cm
(b) Given:
From fig (2)
area of ∆ABD is 24 sq. units. AB = 8 units
To find:
Height of ABC
Now, let us find
We know that ar ∆ABD = 24 sq. units …… (1)
So, ar ∆ABD = ∆ABC ….. (2)
From (1) and (2)
ar ∆ABC = 24 sq. units
½ × AB × height = 24
½ × 8 × height = 24
4 × height = 24
Height = 24/4
= 6
∴ Height of ∆ABC = 6 sq. units
(c) Given:
From fig (3)
In ||gm ABCD, E, and F are midpoints of sides AB and CD respectively.
ar (||gm ABCD) = 36 \mathrm{cm}^{2}
From (1) and (2)
ar (∆ APD) = ½ × 36
= 18 \mathrm{cm}^{2}
(ii) we know that E and F are mid-points of AB and CD
In ∆CPD, EF || PC
Also, EF bisects the ||gm ABCD in two equal parts.
So, EF || AD and AE || DF
AEFD is a parallelogram.
ar (||gm AEFD) = ½ ar (||gm ABCD) ……. (3)
From (1) and (3)
ar (∆APD) = ar (||gm AEFD)
∴ AEFD is the required parallelogram which is equal to the area of ∆APD.
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