Question : A string of length 24 cm is bent first into a square and then into a right-angled by keeping one side of the square fixed as its base. Then the area of a triangle equals to:
Option 1: 24 cm2
Option 2: 60 cm2
Option 3: 40 cm2
Option 4: 28 cm2
Correct Answer: 24 cm 2
Solution : The side of the square = $\frac{24}{4}$ = 6 cm The base of the triangle is 6 cm. The perimeter of the triangle is 24 cm. Using the Pythagoras triplets (i.e. 6, 8, and 10). So, the height of the triangle is 8 cm. The area of a triangle = $\frac{1}{2}$ × 6 cm × 8 cm Hence, the correct answer is 24 cm 2 .
Result | Eligibility | Application | Selection Process | Cutoff | Admit Card | Preparation Tips
Question : The base of a triangle is equal to the perimeter of a square whose diagonal is $6 \sqrt{2}$ cm, and its height is equal to the side of a square whose area is 144 cm2. The area of the triangle (in cm2) is:
Question : The base of a triangle is equal to the perimeter of a square whose diagonal is $9 \sqrt{2}$ cm and its height is equal to the side of a square whose area is 144 cm2. The area of the triangle(in cm2) is:
Question : The total surface area of a right pyramid on a square base of side 10 cm with height 12 cm is:
Question : A right-angled isosceles triangle is inscribed in a semi-circle of radius 7 cm. The area enclosed by the semi-circle but exterior to the triangle is:
Question : A right square pyramid having a lateral surface area is 624 cm2. If the length of the diagonal of the square is $24 \sqrt{2}$, then the volume of the pyramid is:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile