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

Answer: (a)

Breadth of the rectangle is 8 units, as solved above

Answer: (d)

In ΔABC and ΔPQC,

∴ ${PC}/{AC} = {PQ}/{AB}$

⇒ $b/{c + a} = a/x$

∴ x = ${a(c + b)}/b = {ac}/b + a$

Answer: (c)

The Given,

⇒ ∠EDC = ∠ECD = 35°

Since, ∠OCD = 55°

Then, ∠OCE = 20°

By using then theorem that triangle on the same segment of a circle makes as equal angles.

Here, OE is a segment, which makes a ΔOFE and ΔOCE.

Therefore, ∠OCE = ∠EFO = 20°

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: (c)

Remaining perimeter

= $({2 πr}/4)$4 + 10 + 2 + 10 + 2 = 2 × 3.14 × 4 + 24

= 25.12 + 24 = 49.12 cm

= 49.1 cm (approx)

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°