CBSE Class 10 Sample Paper for 2020 Boards - Maths Basic

Question 29 - CBSE Class 10 Sample Paper for 2020 Boards - Maths Basic - Solutions of Sample Papers for Class 10 Boards

Last updated at Oct. 24, 2019 by Teachoo

Following figure depicts a park where two opposite sides are parallel and left and right ends are semi-circular in shape. It has a 7 m wide track for walking

Two friends Seema and Meena went to the park. Meena said that area of the track is 4066 m
^{
2
}
. Is she right? Explain.

Note
: This
is similar
to Ex 12.3, 8 of NCERT – Chapter 12 Class 10

Question 29 Following figure depicts a park where two opposite sides are parallel and left and right ends are semi-circular in shape. It has a 7 m wide track for walking Two friends Seema and Meena went to the park. Meena said that area of the track is 4066 m2. Is she right? Explain.
Let us label the figure
Area of track
= Area ABFE + Area HGCD
+ (Area of semicircle with diameter AD
– Area of semicircle with diameter EH)
+ (Area of semicircle with diameter BC
– Area of semicircle with diameter FG)
Area ABFE
ABFE is a rectangle
with length = 120 m
& breadth = 7 m
Area of ABFE = Length × Breadth
= 120 × 7
= 840 m2
Area HGCD
HGCD is a rectangle
with length = 120 m
& breadth = 7 m
Area of HGCD = Length × Breadth
= 120 × 7
= 840 m2
Area semicircle with diameter AD
Diameter = AD = 70 m
Radius = r = 70/2 = 35 m
Area semicircle AOD = 1/2×𝜋𝑟2
= 1/2×22/7×35×35
= 1925 m2
Area semicircle with diameter EH
Diameter = EH = 70 – 7 – 7 = 56 m
Radius = r = 56/2 = 28 m
Area semicircle EOH = 1/2×𝜋𝑟2
= 1/2×22/7×28×28
= 1232 m2
Area semicircle with diameter BC
Diameter = BC = 70 m
Radius = r = 70/2 = 35 m
Area semicircle BO’C = 1/2×𝜋𝑟2
= 1/2×22/7×35×35
= 1925 m2
Area semicircle with diameter FG
Diameter = FG = 70 – 7 – 7 = 56 m
Radius = r = 56/2 = 28 m
Area semicircle FO’G = 1/2×𝜋𝑟2
= 1/2×22/7×28×28
= 1232 m2
Area of track
= Area ABFE + Area HGCD
+ (Area of semicircle with diameter AD
– Area of semicircle with diameter EH)
+ (Area of semicircle with diameter BC
– Area of semicircle with diameter FG)
= 840 + 840 + (1925 – 1232) + (1925 – 1232)
= 2 × 840 + 2 (1925 – 1232)
= 1680 + 2 (693)
= 1680 + 1386
= 3066 m2
Since area of track is not 4066 m2
∴ Meena was wrong

CBSE Class 10 Sample Paper for 2020 Boards - Maths Basic

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.