SSC CGL Tier-1 practice question
Find the area of a triangle with base 4 cm and height 9 cm.
From a SSC CGL Tier-1 Quantitative Aptitude practice set · Quantitative Aptitude
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Correct answer: (B) 18 cm²
Given:
$$\text{Base } (b) = 4 \text{ cm}$$
$$\text{Height } (h) = 9 \text{ cm}$$
$$\text{Area of triangle} = \frac{1}{2} \times b \times h = \frac{1}{2} \times 4 \times 9 = 2 \times 9 = 18 \text{ cm}^2$$
$$\text{Base } (b) = 4 \text{ cm}$$
$$\text{Height } (h) = 9 \text{ cm}$$
$$\text{Area of triangle} = \frac{1}{2} \times b \times h = \frac{1}{2} \times 4 \times 9 = 2 \times 9 = 18 \text{ cm}^2$$
हिंदी में प्रश्न
4 सेमी आधार और 9 सेमी ऊँचाई वाले त्रिभुज का क्षेत्रफल ज्ञात करें।
उत्तर जाँचने के लिए किसी विकल्प पर टैप करें।
उत्तर और व्याख्या देखें
सही उत्तर: (B) 18 cm²
दिया गया है:
$$\text{आधार } (b) = 4 \text{ cm}$$
$$\text{ऊँचाई } (h) = 9 \text{ cm}$$
$$\text{त्रिभुज का क्षेत्रफल} = \frac{1}{2} \times b \times h = \frac{1}{2} \times 4 \times 9 = 2 \times 9 = 18 \text{ cm}^2$$
$$\text{आधार } (b) = 4 \text{ cm}$$
$$\text{ऊँचाई } (h) = 9 \text{ cm}$$
$$\text{त्रिभुज का क्षेत्रफल} = \frac{1}{2} \times b \times h = \frac{1}{2} \times 4 \times 9 = 2 \times 9 = 18 \text{ cm}^2$$
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