Case Study Questions Class 7 Maths – Mensuration
CBSE Class 7 Case Study Questions Maths Mensuration. Here we have arranged some Important Case Base Questions for students who are searching for Paragraph Based Questions Mensuration.
Directions (Questions 1 to 4): Examine the following case study carefully and answer the questions that follow.
Bimal has a flexible string which is 132 cm long. He bent it into different shapes one by one.
(1) Bimal bent the string to form an equilateral triangle. What will be the length of each side of the triangle formed?
(a) 11 cm
(b) 22 cm
(c) 44 cm
(d) 48 cm
Ans: (c) ⇒ 3 × length of each side = 132 cm ⇒ length of each side = 132/3 cm = 44 cm.
(2) Bimal bent the string to form a square. Find the area of the square.
(a) 1,936 cm2.
(b) 1,729 cm2.
(c) 1,464 cm2.
(d) 1.089 cm2.
Ans: (d) Perimeter of square = 132 cm.
⇒ 4 × side = 132 cm ⇒ side = 1328/4 cm = 33 cm.
⇒area of square = (side)2 = (33)2 cm2 = 1,089 cm2.
(3) He bent it into a circle. Find the radius of the circle formed. [Use = π = 22/7]
(a) 7.5 cm
(b) 10.5 cm
(c) 15 cm
(d) 21 cm
Ans: (d) Circumference of the circle = 132 cm.
⇒ 2πr = 132 cm ⇒ 2× 22/7 × r = 132 cm ⇒ r = (132 × 7/2 × 22) = 21 cm.
∴radius of the circle = 21 cm.
(4) The area of the circle formed by bending the string will be [Use π = 22/7].
(a) 1,323cm2
(b) 1,386 cm2
(c) 1,545 cm2
(d) 1,594 cm2
Ans: (b) Area of the circle = πr2 = (22/7 × 21 × 21) cm2 = 1.386 cm2.
Directions (Questions 1 to 4): Examine the following case study carefully and answer the questions that follow.
Krishna has a farmland in the shape of a rhombus PQRS as shown in the adjoining figure. The length of each side of this land is 450 m. Krishna fixed a wire PM such that PM ⊥QR and another wire along the diagonal PR. He measured PM and PR and found that PM = 75 m and PR = 125 m. He grew potatoes in the triangular region △PMR and sugarcane in the triangular region △PSR. The remaining area was left for cattle rearing.
(1) The area of the farmland PQRS is –
(a) 33,750 m2
(b) 56,250 m2
(c) 17,525 m2
(d) 9,375 m2
Ans: (a) Area of farmland PQRS (rhombus) = base × height = QR × PM
= 450 m × 75 m = 33,750 m2.
(2) The length of the diagonal QS of the rhombus PQRS is –
(a) 360 m
(b) 450 m
(c) 480 m
(d) 540 m
Ans: Area of rhombus = ½ × product of diagonals
⇒ 33,750 = ½ × PR × QS ⇒ 33,750 = ½ × 125 × QS ⇒ QS = 2 × 33,750/125 = 540 m.
(3) The area of the field used for growing potatoes is –
(a) 3,750m2
(b) 4,225 m2
(c) 6,250 m2
(d) 7,500 m2
Ans: (a) In right △PMR,
MR = √PR2 – PM2 = √(125)2 – (75)2 M = √15,625 – 5,625 m = √10,000 m = 100m.
Now, Area (right △PMR) = ½ × MR × pm = (1/2 × 100 × 75) m2 = 3,750 m2.
∴area of field (△PMR) used for growing potatoes = 3,750 m2.
(4) The area of the field used for growing sugarcane is –
(a) 17,525m2
(b) 9,375 m2
(c) 16,875m2
(d) 10,250 m2
Ans: (c) Area (△PQR) = ½ × QR × PM = (1/2 × 450 × 75) m2 = 16,875 m2.
∴ area (△PSR) = area (rhombus PQRS) – area (△PQR)
= (33,750 – 16,875) m2 = 16,875 m2.
Thus, the area of the field (△PSR) used for growing sugarcane = 16,875 m2.