Mensuration Model Questions Set 2 Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (a)

In ΔABC, we draw a line l || BF which intersect AC at G. In ΔADG and ΔAEF;

given that EA is the mid point of AD and DL || EF.

So, concept of similar triangle.

F is also mid point of AG. AF = FG (i)

DADG and DAEF are similar.

Again

ΔFBC and ΔDCl

BF || DG

given that AD is median so thad DB the mid point of BC.

G will be The mid point of CF

CG = GF ...(ii)

From equations (i) and (ii), we get

AF = FG = CG ...(iii)

From figure, AC = AF = FG + CG

= AF + AF + AF + 3AF

⇒ AF = $1/3$ AC

Answer: (b)

length of the are = 2 π r $(θ/{360º})$

Radius of arc(r) = 30 am

Length of the arc = 2πr = 2π × 30 × ${144}/{360}$ = 24π

Let the radius of the cone = R ∴ 2πR = 24π

⇒ R = ${30 × 144}/{360}$ = 12cm

Answer: (a)

Breadth of the rectangle is 8 units, as solved above

Answer: (c)

Given, AC = BC

In ΔABC, ∠ABC = ∠CAB

∠ABC = ∠CAB = 38° (∴ AC = BC)

∠ACB = 180° – (∠ABC + ∠CAB)

= 180° – (38° + 38°) = 180° – 76° = 104°

In ΔACD, ∠ACD = 180° – 104° = 76°

and ∠ACD = ∠CAD = 76°

(∵ CD = AD)

∴ ∠ADC = 180° – (∠ACD + ∠CAD)

= 180° – (76° + 76°) = 28°

Answer: (c)

Area of parallelogram = 6 × Area of ΔNPR

∴ NR × PL= 6 × $1/2$ × NR × PR

⇒ PL = 3PR (here, PL = PR + RL)

⇒ PR + RL= 3PR

⇒ RL= 2PL = 2 × 6 = 12 cm

Answer: (d)

Let O is centre of big circle, and O' is centre of smaller circle. Both are touch internally each other.

OA = 6 cm O'A = 4cm

Here PR is longest chord of big circle

P.S = ${PR}/2$

OS = AS – OA

= 8 – 6 = 2 cm.

In ΔPSO

$(PS)^2 + (OS)^2 = (OP)^2$

⇒ $(PS)^2 + (2)^2 = (6)^2$

PS = $√{36 - 4} = √{32} = 4 √2$

Now,

PS = ${PR}/2$

PR = 2 × PS

= 2 × $4 √2 = 8 √2$