The base of a right prism is a quadrilateral | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) In right $\Delta DAB$,by the Pythagorean theorem,$BD = \sqrt{AD^2 + AB^2} = \sqrt{12^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225} = 15\, cm$.Area of $\

Vedclass · Accepted Answer

$720$

Vedclass · Answer

(A) In right $\Delta DAB$,by the Pythagorean theorem,$BD = \sqrt{AD^2 + AB^2} = \sqrt{12^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225} = 15\, cm$.Area of $\Delta DAB = \frac{1}{2} 	imes AB 	imes AD = \frac{1}{2} 	imes 9 	imes 12 = 54\, cm^2$.For $\Delta BCD$,the sides are $BC = 14\, cm$,$CD = 13\, cm$,and $BD = 15\, cm$. The semi-perimeter $S = \frac{14 + 13 + 15}{2} = \frac{42}{2} = 21\, cm$.Using Heron's formula,Area of $\Delta BCD = \sqrt{S(S-a)(S-b)(S-c)} = \sqrt{21(21-14)(21-13)(21-15)} = \sqrt{21 	imes 7 	imes 8 	imes 6} = \sqrt{7056} = 84\, cm^2$.Area of quadrilateral $ABCD = 	ext{Area of } \Delta DAB + 	ext{Area of } \Delta BCD = 54 + 84 = 138\, cm^2$.Volume of the prism = Base Area $	imes$ Height.$2070 = 138 	imes h \implies h = \frac{2070}{138} = 15\, cm$.Lateral Surface Area $(LSA)$ = Perimeter of base $	imes$ Height.Perimeter = $AB + BC + CD + DA = 9 + 14 + 13 + 12 = 48\, cm$.$LSA$ = $48 	imes 15 = 720\, cm^2$.

The base of a right prism is a quadrilateral $ABCD$. Given that $AB = 9\, cm$,$BC = 14\, cm$,$CD = 13\, cm$,$DA = 12\, cm$ and $\angle DAB = 90^{\circ}$. If the volume of the prism is $2070\, cm^3$,then the area of the lateral surface is........$cm^2$.

Similar Questions

The cross section of a canal is in the shape of an isosceles trapezium which is $4 \ m$ wide at the bottom and $5 \ m$ wide at the top. If the depth of the canal is $2 \ m$ and it is $120 \ m$ long,what is the maximum capacity of this canal? (in $m^3$)

The volumes of a right circular cylinder and a sphere are equal. The radius of the cylinder and the diameter of the sphere are equal. The ratio of the height to the radius of the cylinder is:

$A$ rectangular water tank is $80 \, m \times 40 \, m$. Water flows into it through a pipe with an opening area of $40 \, cm^2$ at a speed of $10 \, km/h$. By how much will the water level in the tank rise in half an hour (in $cm$)?

The radius of a wheel is $21 \, cm$. What is the distance (in $cm$) travelled by the wheel in $10$ revolutions?

$2$ cubes have volumes in the ratio $1:27$. The ratio of the area of the face of one to that of the other is

Vedclass Test Series

Exam Paper Generator

Online Exam Module