Test Level 1: Geometry - 2 - CAT MCQ

When the length of stick = 20 m, then length of shadow = 10 m i.e. in this case length = 2 x shadow.
With the same angle of inclination of the sun, the length of tower that casts a shadow of 50 m ⇒ 2 x 50 m = 100 m
i.e. height of tower = 100 m

For similar triangles fi (Ratio of sides)² = Ratio of areas

Then as per question

{Let the longest side of ΔDEF = x}

Ratio of corresponding sides 

∠x = 35°; because angles subtended by an arc, anywhere on the circumference are equal.

By the rule of tangents, we get
12² = (x + 7) x ⇒ 144 = x² + 7x
⇒ x² + 7x – 144 = 0
⇒ x² + 16x – 9x – 144 = 0
⇒ x(x + 16) – 9(x + 16)
⇒ x = 9 or –16
–16 can’t be the length, hence this value is discarded, thus, x = 9

ADBC is a cyclic quadrilateral as all its four vertices are on the circumference of the circle. Also, the opposite angles of the cyclic quadrilateral are supplementary.
Therefore, ∠ADB = 180 – 48° = 132°

OB = OA = radius of the circle
∠AOB = 180 – (30 + 30)
{Sum of angles of a triangle = 180°)
⇒ 120°
Then

because the angle subtended
by a chord at the centre is twice of what it
can subtend at the circumference. Again, ABCD is a cyclic quadrilateral;
So ∠ACB = 180° – 60° = 120° (because opposite angles of cyclic quadrilateral are supplementary).

Since the lines AB and CD are parallel to each other, and the lines RD and AN are parallel, it means that the triangles RBF and NCI are similar to each other.
Since the ratio of CN:BR = 1.333, if we take BR as 3, we will get CN as 4. This means that the ratio of BF:CI would also be 3:4. Also, the ratio of BR:
RS:ST:TA = BF:FG:GH:HI = 3:5:2:7 (given). Hence, the correct answer is 3:5:2:7:4.

(Ratio of corresponding sides)2 = Ratio of area of similar triangles
∴ Ratio of corresponding sides in this question

BC = ED = 6 m
So AB = AC – BC = 11 – 6 = 5 m
CD = BE = 12 m
Then by Pythagoras theorem:
AE² = AB² + BE² ⇒ AE = 13 m

In the DOBC, OB = 12 cm, OC = radius = 13 cm.
Then using Pythagoras theorem;
BC² = OC² – OB² = 25; BC = 5 cm
Length of the chord = 2 x BC = 2 x 5 = 10 cm

∠AOM = 2∠ABM and
∠AON = 2∠ACN
because angle subtended by an arc at the centre of the circle is twice the angle subtended by it on the circumference on the same segment.
∠AON = 60° and ∠ AOM = 40°
∠X = ∠AON + ∠AOM
(∵ vertically opposite angles).
∠ X = 100°
Alternately, you could also solve this using the following process:

In the given figure, join the points BD and CD. Then, in the cyclic quadrilateral ABDC, the sum of angles x/2 and y would be 180o. Hence, y = 180 – x/2. Also, the sum of the angles OBD + OCD = 180 – 20 - 30
= 130°. Therefore, x + y = 230 (as the sum of the angles of the quadrilateral OBDC is 360). Solving, the two equations, we get x = 100.

By the rule of tangents, we know:
6² = (5 + x)5 ⇒ 36 = 25 + 5x ⇒ 11 = 5x ⇒ x = 2.2 cm

By the rule of chords, cutting externally, we get
⇒ (9 + 6)6 = (5 + x)5 ⇒ 90 = 25 + 5x ⇒ 5x = 65

⇒ x = 13 cm

∠APB = 90° (angle in a semicircle = 90°)
∠PBA = 180 – (90 + 25) = 65°
∠TPA = ∠PBA (the angle that a chord makes
with the tangent, is subtended by the chord on the circumference in the alternate segment).
= 65°

Note: This is also called as the Alternate Segment Theorem.

From the given figure we have:
4b + 2c = 180    (1)
a + b = 105    (2)
4b = a    (3)
Solving these equations, we get that b = 21°; a = 84°; c = 48°.

(because the angle subtended
by an arc on the circumference is half of what it subtends at the centre). ABCD one cyclic quadrilateral
So ∠ABC = 180° – 70° = 110° (because opposite.angles of a cyclic quadrilateral are supplementary).

∠BAC = 30° (∵ angles subtended by an arc anywhere on the circumference in the same segment are equal).

In DBAC; ∠x = 180° – (110° + 30°) = 40°
(∵ sum of angles of a triangle = 180°)

Perimeter of the figure = 10 + 10 + 6 + 6π = 26 + 6π