RS AGGARWAL CLASS 9 CHAPTER 14 AREAS OF TRIANGLES ,QUADRILATERAL MCQ

 MULTIPLE CHOICE QUESTIONS

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Question 1:

In a ∆ABC it is given that base = 12 cm and height = 5 cm. Its area is
(a) 60 cm²
(b) 30 cm²
(c) $15\sqrt{3}{\mathrm{cm}}^2$$15\sqrt{3} {\mathrm{cm}}^2$
(d) 45 cm²

Answer 1:

(b) 30 cm²

$\mathrm{Area}\mathrm{of}\mathrm{triangle}=\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height}\mathrm{Area}\mathrm{of}\Delta ABC=\frac{1}{2}\times 12\times 5=30{\mathrm{cm}}^2$$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{triangle} = \frac{1}{2}\times \mathrm{Base}\times \mathrm{Height} \\ \mathrm{Area} \mathrm{of}\mathit{ }\mathit{∆}ABC=\frac{1}{2}\times 12\times 5=30 {\mathrm{cm}}^2 \end{gathered}$$

Question 2:

The lengths of three sides of a triangle are 20 cm, 16 cm and 12 cm. The area of the triangle is
(a) 96 cm²
(b) 120 cm²
(c) 144 cm²
(d) 160 cm²

Answer 2:

(a) 96 cm²

$\mathrm{Let}:a=20\mathrm{cm},b=16\mathrm{cm}\mathrm{and}c=12\mathrm{cm}s=\frac{a+b+c}{2}=\frac{20+16+12}{2}=24\mathrm{cm}\mathrm{By}\mathrm{Heron}\text{'}s\mathrm{formula},\mathrm{we}\mathrm{have}:\mathrm{Area}\mathrm{of}\mathrm{triangle}=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{24(24-20)(24-16)(24-12)}=\sqrt{24\times 4\times 8\times 12}=\sqrt{6\times 4\times 4\times 4\times 4\times 6}=6\times 4\times 4=96{\mathrm{cm}}^2$$$\begin{gathered} \mathrm{Let}: \\ a=20 \mathrm{cm}, b = 16 \mathrm{cm} \mathrm{and} c=12 \mathrm{cm} \\ s= \frac{a+b+c}{2}=\frac{20+16+12}{2}=24 \mathrm{cm} \\ \mathrm{By} \mathrm{Heron}\text{'}\mathrm{s} \mathrm{formula}, \mathrm{we} \mathrm{have}: \\ \mathrm{Area} \mathrm{of} \mathrm{triangle} = \sqrt{s(s-a)(s-b)(s-c)} \\ =\sqrt{24(24-20)(24-16)(24-12)} \\ =\sqrt{24\times 4\times 8\times 12} \\ =\sqrt{6\times 4\times 4\times 4\times 4\times 6} \\ =6\times 4\times 4 \\ =96 {\mathrm{cm}}^2 \end{gathered}$$

Question 3:

Each side of an equilateral triangle measure 8 cm. The area of the triangle is
(a) $8\sqrt{3}{\mathrm{cm}}^2$$8\sqrt{3} {\mathrm{cm}}^2$
(b) $16\sqrt{3}{\mathrm{cm}}^2$$16\sqrt{3} {\mathrm{cm}}^2$
(c) $32\sqrt{3}{\mathrm{cm}}^2$$32\sqrt{3} {\mathrm{cm}}^2$
(d) 48 cm²

Answer 3:

(b) $16\sqrt{3}{\mathrm{cm}}^2$$16\sqrt{3} {\mathrm{cm}}^2$
$\mathrm{Area}\mathrm{of}\mathrm{equilateral}\mathrm{triangle}=\frac{\sqrt{3}}{4}\times (\mathrm{Side}{)}^2=\frac{\sqrt{3}}{4}\times (8{)}^2=\frac{\sqrt{3}}{4}\times 64=16\sqrt{3}{\mathrm{cm}}^2$$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{equilateral} \mathrm{triangle} = \frac{\sqrt{3}}{4}\times (\mathrm{Side}{)}^2 \\ =\frac{\sqrt{3}}{4}\times (8{)}^2 \\ =\frac{\sqrt{3}}{4}\times 64 \\ =16\sqrt{3} {\mathrm{cm}}^2 \end{gathered}$$

Question 4:

The base of an isosceles triangle is 8 cm long and each of its equal sides measures 6 cm. The area of the triangle is
(a) $16\sqrt{5}{\mathrm{cm}}^2$$16\sqrt{5} {\mathrm{cm}}^2$
(b) $8\sqrt{5}{\mathrm{cm}}^2$$8\sqrt{5} {\mathrm{cm}}^2$
(c) $16\sqrt{3}{\mathrm{cm}}^2$$16\sqrt{3} {\mathrm{cm}}^2$
(d) $8\sqrt{3}{\mathrm{cm}}^2$$8\sqrt{3} {\mathrm{cm}}^2$

Answer 4:

(b) $8\sqrt{5}{\mathrm{cm}}^2$$8\sqrt{5} {\mathrm{cm}}^2$
$\mathrm{Area}\mathrm{of}\mathrm{isosceles}\mathrm{triangle}=\frac{b}{4}\sqrt{4a^2-b^2}\mathrm{Here},a=6\mathrm{cm}\mathrm{and}b=8\mathrm{cm}\mathrm{Thus},\mathrm{we}\mathrm{have}:\frac{8}{4}\times \sqrt{4(6{)}^2-8^2}=\frac{8}{4}\times \sqrt{144-64}=\frac{8}{4}\times \sqrt{80}=\frac{8}{4}\times 4\sqrt{5}=8\sqrt{5}{\mathrm{cm}}^2$$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{isosceles} \mathrm{triangle} = \frac{b}{4}\sqrt{4a^2-b^2} \\ \mathrm{Here}, \\ a= 6 \mathrm{cm} \mathrm{and} b=8 \mathrm{cm} \\ \mathrm{Thus}, \mathrm{we} \mathrm{have}: \\ \frac{8}{4}\times \sqrt{4(6{)}^2-8^2} \\ =\frac{8}{4}\times \sqrt{144-64} \\ =\frac{8}{4}\times \sqrt{80} \\ =\frac{8}{4}\times 4\sqrt{5} \\ =8\sqrt{5} {\mathrm{cm}}^2 \end{gathered}$$

Question 5:

The base of an isosceles triangle is 6 cm and each of its equal sides is 5 cm. The height of the triangle is
(a) 8 cm
(b) $\sqrt{30}\mathrm{cm}$$\sqrt{30} \mathrm{cm}$
(c) 4 cm
(d) $\sqrt{11}\mathrm{cm}$$\sqrt{11} \mathrm{cm}$

Answer 5:

(c) 4 cm
$\mathrm{Height}\mathrm{of}\mathrm{isosceles}\mathrm{triangle}=\frac{1}{2}\sqrt{4a^2-b^2}=\frac{1}{2}\sqrt{4{\left(5\right)}^2-6^2}\left(a=5\mathrm{cm}\mathrm{and}b=6\mathrm{cm}\right)=\frac{1}{2}\times \sqrt{100-36}=\frac{1}{2}\times \sqrt{64}=\frac{1}{2}\times 8=4\mathrm{cm}$$$\begin{gathered} \mathrm{Height} \mathrm{of} \mathrm{isosceles} \mathrm{triangle} = \frac{1}{2}\sqrt{4a^2-b^2} \\ =\frac{1}{2}\sqrt{4{\left(5\right)}^2-6^2} \left(a=5 \mathrm{cm} \mathrm{and} b=6 \mathrm{cm}\right) \\ =\frac{1}{2}\times \sqrt{100-36} \\ =\frac{1}{2}\times \sqrt{64} \\ =\frac{1}{2}\times 8 \\ =4 \mathrm{cm} \end{gathered}$$

Question 6:

Each of the two equal sides of an isosceles right triangle is 10 cm long. Its area is
(a) $5\sqrt{10}{\mathrm{cm}}^2$$5\sqrt{10} {\mathrm{cm}}^2$
(b) 50 cm²
(c) $10\sqrt{3}{\mathrm{cm}}^2$$10\sqrt{3} {\mathrm{cm}}^2$
(d) 75 cm²

Answer 6:

(b) 50 cm²
Here, the base and height of the triangle are 10 cm and 10 cm, respectively.
Thus, we have:
$\mathrm{Area}\mathrm{of}\mathrm{triangle}=\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height}=\frac{1}{2}\times 10\times 10=50{\mathrm{cm}}^2$$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{triangle} = \frac{1}{2}\times \mathrm{Base}\times \mathrm{Height} \\ =\frac{1}{2}\times 10\times 10 \\ =50 {\mathrm{cm}}^2 \end{gathered}$$

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Question 7:

Each side of an equilateral triangle is 10 cm long. The height of the triangle is
(a) $10\sqrt{3} \mathrm{cm}$
(b) $5\sqrt{3} \mathrm{cm}$
(c) $10\sqrt{2} \mathrm{cm}$
(d) 5 cm

Answer 7:

(b) $5\sqrt{3} \mathrm{cm}$
$$\begin{gathered} \mathrm{Height} \mathrm{of} \mathrm{equilateral} \mathrm{triangle}=\frac{\sqrt{3}}{2}\times \mathrm{Side} \\ =\frac{\sqrt{3}}{2}\times 10 \\ =5\sqrt{3} \mathrm{cm} \end{gathered}$$

Question 8:

The height of an equilateral triangle is 6 cm. Its area is
(a) $12\sqrt{3} {\mathrm{cm}}^2$
(b) $6\sqrt{3} {\mathrm{cm}}^2$
(c) $12\sqrt{2} {\mathrm{cm}}^2$
(d) 18 cm²

Answer 8:

(a) $12\sqrt{3} {\mathrm{cm}}^2$
$$\begin{gathered} \mathrm{Height} \mathrm{of} \mathrm{equilateral} \mathrm{triangle} = \frac{\sqrt{3}}{2}\times \mathrm{Side} \\ \Rightarrow 6=\frac{\sqrt{3}}{2}\times \mathrm{Side} \\ \Rightarrow \mathrm{Side}=\frac{12}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{12}{3}\times \sqrt{3}=4\sqrt{3 } \mathrm{cm} \\ \mathrm{Now}, \\ \mathrm{Area} \mathrm{of} \mathrm{equilateral} \mathrm{triangle} = \frac{\sqrt{3}}{4}\times (\mathrm{Side}{)}^2 \\ =\frac{\sqrt{3}}{4}\times {\left(4\sqrt{3}\right)}^2 \\ =\frac{\sqrt{3}}{4}\times 48 \\ =12\sqrt{3} {\mathrm{cm}}^2 \end{gathered}$$

Question 9:

The lengths of the three sides of a triangular field are 40 m, 24 m and 32 m respectively. The area of the triangle is
(a) 480 m²
(b) 320 m²
(c) 384 m²
(d) 360 m²

Answer 9:

(c) 384 m²

$$\begin{gathered} \mathrm{Let}: \\ a=40 \mathrm{m}, b = 24 \mathrm{m} \mathrm{and} c=32 \mathrm{m} \\ s= \frac{a+b+c}{2}=\frac{40+24+32}{2}=48 \mathrm{m} \\ \mathrm{By} \mathrm{Heron}\text{'}\mathrm{s} \mathrm{formula}, \mathrm{we} \mathrm{have}: \\ \mathrm{Area} \mathrm{of} \mathrm{triangle} = \sqrt{s(s-a)(s-b)(s-c)} \\ =\sqrt{48(48-40)(48-24)(48-32)} \\ =\sqrt{48\times 8\times 24\times 16} \\ =\sqrt{24\times 2\times 8\times 24\times 8\times 2} \\ =24\times 8\times 2 \\ =384 {\mathrm{m}}^2 \end{gathered}$$

Question 10:

The sides of a triangle are in the ratio 5 : 12 : 13 and its perimeter is 150 cm. The area of the triangle is
(a) 375 cm²
(b) 750 cm²
(c) 250 cm²
(d) 500 cm²

Answer 10:

(b) 750 cm²

Let the sides of the triangle be 5x cm, 12x cm and 13x cm.
Perimeter = Sum of all sides
or, 150 = 5x + 12x + 13x
or, 30x = 150
or, x = 5
Thus, the sides of the triangle are 5$\times$5 cm, 12$\times$5 cm and 13$\times$5 cm, i.e., 25 cm, 60 cm and 65 cm.

Now,
$$\begin{gathered} \mathrm{Let}: \\ a=25 \mathrm{cm}, b = 60 \mathrm{cm} \mathrm{and} c=65 \mathrm{cm} \\ s= \frac{150}{2}=75 \mathrm{cm} \\ \mathrm{By} \mathrm{Heron}\text{'}\mathrm{s} \mathrm{formula}, \mathrm{we} \mathrm{have}: \\ \mathrm{Area} \mathrm{of} \mathrm{triangle} = \sqrt{s(s-a)(s-b)(s-c)} \\ =\sqrt{75(75-25)(75-60)(75-65)} \\ =\sqrt{75\times 50\times 15\times 10} \\ =\sqrt{15\times 5\times 5\times 10\times 15\times 10} \\ =15\times 5\times 10 \\ =750 {\mathrm{cm}}^2 \end{gathered}$$

Question 11:

The lengths of the three sides of a triangle are 30 cm, 24 cm and 18 cm respectively. The length of the altitude of the triangle corresponding to the smallest side is
(a) 24 cm
(b) 18 cm
(c) 30 cm
(d) 12 cm

Answer 11:

(a) 24 cm

$$\begin{gathered} \mathrm{Let}: \\ a=30 \mathrm{cm}, b = 24 \mathrm{cm} \mathrm{and} c=18 \mathrm{cm} \\ s= \frac{a+b+c}{2}=\frac{30+24+18}{2}=36 \mathrm{cm} \\ \mathrm{On} \mathrm{applying} \mathrm{Heron}\text{'}\mathrm{s} \mathrm{formula}, \mathrm{we} \mathrm{get}: \\ \mathrm{Area} \mathrm{of} \mathrm{triangle} = \sqrt{s(s-a)(s-b)(s-c)} \\ =\sqrt{36(36-30)(36-24)(36-18)} \\ =\sqrt{36\times 6\times 12\times 18} \\ =\sqrt{12\times 3\times 12\times 6\times 3} \\ =12\times 3\times 6 \\ =216 {\mathrm{cm}}^2 \end{gathered}$$

The smallest side is 18 cm.
Hence, the altitude of the triangle corresponding to 18 cm is given by:
$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{triangle} = 216 {\mathrm{cm}}^2 \\ \Rightarrow \frac{1}{2}\times \mathrm{Base}\times \mathrm{Height} = 216 \\ \Rightarrow \mathrm{Height} = \frac{216\times 2}{18}=24 \mathrm{cm} \end{gathered}$$

Question 12:

The base of an isosceles triangle is 16 cm and its area is 48 cm². The perimeter of the triangle is
(a) 41 cm
(b) 36 cm
(c) 48 cm
(d) 324 cm

Answer 12:

(b) 36 cm

Let $△$PQR be an isosceles triangle and PX$\perp$QR.
Now,
$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{triangle} =48 {\mathrm{cm}}^2 \\ \Rightarrow \frac{1}{2}\times QR\times PX = 48 \\ \Rightarrow h =\frac{96}{16}=6 \mathrm{cm} \\ \mathrm{Also},\mathit{ } \\ QX\mathit{ }= \frac{1}{2}\times 24 = 12 \mathrm{cm} \mathrm{and} PX = 12 \mathrm{cm} \end{gathered}$$
$$\begin{gathered} PQ=\sqrt{Q{X}^2+P{X}^2} \\ a=\sqrt{8^2+6^2}=\sqrt{64+36}=\sqrt{100}=10 \mathrm{cm} \end{gathered}$$

∴ Perimeter = (10 + 10 + 16) cm = 36 cm

Question 13:

The area of an equilateral triangle is $36\sqrt{3} {\mathrm{cm}}^2$. Its perimeter is
(a) 36 cm
(b) $12\sqrt{3} \mathrm{cm}$
(c) 24 cm
(d) 30 cm

Answer 13:

(a) 36 cm
$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{equilateral} \mathrm{triangle} = \frac{\sqrt{3}}{4}\times (\mathrm{Side}{)}^2 \\ \Rightarrow \frac{\sqrt{3}}{4}\times (\mathrm{Side}{)}^2 =36\sqrt{3} \end{gathered}$$

$$\begin{gathered} \Rightarrow (\mathrm{Side}{)}^2=144 \\ \Rightarrow \mathrm{Side}=12 \mathrm{cm} \end{gathered}$$

Now,
Perimeter = 3 × Side = 3 × 12 = 36 cm

Question 14:

Each of the equal sides of an isosceles triangle is 13 cm and its base is 24 cm. The area of the triangle is
(a) 156 cm²
(b) 78 cm²
(c) 60 cm²
(d) 120 cm²

Answer 14:

(c) 60 cm²
$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{isosceles} \mathrm{triangle} = \frac{b}{4}\sqrt{4a^2-b^2} \\ \mathrm{Here}, \\ a= 13 \mathrm{cm} \mathrm{and} b=24 \mathrm{cm} \\ \mathrm{Thus}, \mathrm{we} \mathrm{have}: \\ \frac{24}{4}\times \sqrt{4(13{)}^2-{24}^2} \\ =6\times \sqrt{676-576} \\ =6\times \sqrt{100} \\ =6\times 10 \\ =60 {\mathrm{cm}}^2 \end{gathered}$$

Question 15:

The base of a right triangle. is 48 cm and its hypotenuse is 50 cm long. The area of the triangle is
(a) 168 cm²
(b) 252 cm²
(c) 336 cm²
(d) 504 cm²

Answer 15:

(c) 336 cm²

Let $△$PQR be a right-angled triangle and PQ$\perp$QR.
Now,
$$\begin{gathered} PQ=\sqrt{P{R}^2-Q{R}^2} \\ =\sqrt{{50}^2-{48}^2} \\ =\sqrt{2500-2304} \\ =\sqrt{196} \\ =14 \mathrm{cm} \end{gathered}$$

$\therefore \mathrm{Area} \mathrm{of} \mathrm{triangle} =\frac{1}{2}\times QR\times PQ =\frac{1}{2}\times 48\times 14=336 {\mathrm{cm}}^2$

Question 16:

The area of an equilateral triangle is $81\sqrt{3} {\mathrm{cm}}^2$. Its height is
(a) $9\sqrt{3} \mathrm{cm}$
(b) $6\sqrt{3} \mathrm{cm}$
(c) $18\sqrt{3} \mathrm{cm}$
(d) 9 cm

Answer 16:

(a) $9\sqrt{3} \mathrm{cm}$
$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{equilateral} \mathrm{triangle} = 81\sqrt{3} {\mathrm{cm}}^2 \\ \Rightarrow \frac{\sqrt{3}}{4}\times (\mathrm{Side}{)}^2=81\sqrt{3} \\ \Rightarrow (\mathrm{Side}{)}^2=81\times 4 \\ \Rightarrow (\mathrm{Side}{)}^2=324 \\ \Rightarrow \mathrm{Side}=18 \mathrm{cm} \\ \mathrm{Now}, \\ \mathrm{Height} = \frac{\sqrt{3}}{2}\times \mathrm{Side}=\frac{\sqrt{3}}{2}\times 18=9\sqrt{3} \mathrm{cm} \end{gathered}$$