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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 figure (1) given below, the perimeter of the parallelogram is 42 cm.
Calculate the lengths of the sides of the parallelogram.
(b) In the figure (2) given below, the perimeter of ∆ ABC is 37 cm. If the lengths of
the altitudes AM, BN and CL are 5x, 6x, and 4x respectively, Calculate the lengths
of the sides of ∆ABC.
(c) In the fig. (3) Given below, ABCD is a parallelogram. P is a point on DC such
that area of ∆DAP = 25 cm² and area of ∆BCP = 15 cm². Find
(i) area of || gm ABCD
(ii) DP: PC.
(a) Given:
The perimeter of parallelogram ABCD = 42 cm
To find:
Lengths of the sides of the parallelogram ABCD.
From fig (1)
We know that,
AB = P
Then, perimeter of ||gm ABCD = 2 (AB + BC)
42 = 2(P + BC)
42/2 = P + BC
21 = P + BC
BC = 21 – P
So, ar (||gm ABCD) = AB × DM
= P × 6
= 6P ………. (1)
Again, ar (||gm ABCD) = BC × DN
= (21 - P) × 8
= 8(21 - P) ………. (2)
From (1) and (2), we get
6P = 8(21 - P)
6P = 168 – 8P
6P + 8P = 168
14P = 168
P = 168/14
= 12
Hence, sides of ||gm are
AB = 12cm and BC = (21 - 12)cm = 9cm
(b) Given:
The perimeter of ∆ ABC is 37 cm. The lengths of the altitudes AM, BN and CL are 5x,
6x, and 4x respectively.
To find:
Lengths of the sides of ∆ABC. i.e., BC, CA, and AB.
Let us consider BC = P and CA = Q
From fig (2),
Then, perimeter of ∆ABC = AB + BC + CA
37 = AB + P + Q
AB = 37 – P – Q
Area (∆ABC) = ½ × base × height
= ½ × BC × AM = ½ × CA × BN = ½ × AB × CL
= ½ × P × 5x = ½ × Q × 6x = ½ (37 – P – Q) × 4x
= 5P/2 = 3Q = 2(37 – P – Q)
Let us consider first two parts:
5P/2 = 3Q
5P = 6Q
5P – 6Q = 0 …… (1)
25P – 30Q (multiplying by 5)….. (2)
Let us consider the second and third parts:
3Q = 2(37 – P – Q)
3Q = 74 – 2P – 2Q
3Q + 2Q + 2P = 74
2P + 5Q = 74 ……. (3)
12P + 30Q = 444 (multiplying by 6)……. (4)
By adding (2) and (4), we get
37P = 444
P = 444/37
= 12
Now, substitute the value of P in equation (1), we get
5P – 6Q = 0
5(12) – 6Q = 0
60 = 6Q
Q = 60/6
= 10
Hence, BC = P = 12cm
CA = Q = 10cm
And AB = 37 – P – Q = 37 – 12 – 10 = 15cm
(c) Given:
ABCD is a parallelogram. P is a point on DC such that area of ∆DAP = 25 cm² and the area
of ∆BCP = 15 cm².
To Find:
(i) area of || gm ABCD
(ii) DP: PC
Now let us find,
From fig (3)
(i) we know that,
ar (∆APB) = ½ ar (||gm ABCD)
Then,
½ ar (||gm ABCD) = ar (∆DAP )
= 25 + 15
\begin{array}{l} =40 \mathrm{cm}^{2} \\ \text { So, ar }(\| \mathrm{gm} \mathrm{ABCD})=2 \times 40=80 \mathrm{cm}^{2} \end{array}
(ii) we know that,
∆ADP and ∆BCP are on the same base CD and between the same parallel lines CD and AB.
ar (∆DAP)/ar(∆BCP) = DP/PC
25/15 = DP/PC
5/3 = DP/PC
So, DP: PC = 5: 3
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