A vertical stick 20 m long casts a shadow 10 m long on the ground. At the same time, a tower casts a shadow 50 m long on the ground. Find the height of the tower.
The area of similar triangles, ABC and DEF are 144cm^{2} and 81 cm^{2} respectively. If the longest side of the larger △ABC be 36 cm, then the longest side of the smaller △DEF is:
For similar triangles
⇒ (Ratio of sides)^{2 } = Ratio of areas
Then as per question = (36/x)^{2 }= 141/81
Let the longest side of ΔDEF = x
⇒ 36/x = 12/9
⇒ x = 27cm
Find the value of x in the figure, if it is given that AC and BD are diameters of the circle.
Find the value of x in the given figure.
By the rule of tangents, we get:
⇒ 12^{2} = (x + 7)x
⇒ 144 = x^{2} + 7x
⇒ x^{2} + 7x – 144 = 0
⇒ x^{2} +16x – 9x –144 = 0
⇒ x(x + 16) – 9(x + 16) = 0
⇒ x = 9 or –16
–16 can’t be the length, hence this value is discarded. Thus, x = 9
Find the value of x in the given figure.
By the rule of chords, cutting externally, we get:
(9 + 6) * 6 = (5 + x) * 5
90 = 25 + 5x
5x = 65
x = 13 cm
In the given figure ㄥPAB = 25º. AB is the diameter. Find ㄥTPA.
Note: This is also called the Alternate Segment Theorem.
In the figure, AB is parallel to CD and RD  SL  TM  AN, and BR : RS : ST : TA = 3 : 5 : 2 : 7. If it is known that CN = 1.333 BR. Find the ratio of BF : FG : GH : HI : IC
Hence, the correct answer is 3 : 5 : 2 : 7 : 4
In the following figure, it is given that O is the centre of the circle and ㄥAOC = 140°. Find ㄥABC.
∠AOC of minor sector = 140°
∠AOC of major sector=360°  140° = 220°
Theorem: The angle subtended at the centre is twice the angle formed at the circumference of the circle.
Hence,
∴ The measure of ∠x = 110°
In the figure below, PQ = QS, QR = RS and angle SRQ = 100°. How many degrees is angle QPS?
In ΔQRS, QR = RS
⇒ ㄥRQS = ㄥRSQ (because angles opposite to equal sides are equal).
Thus:
ㄥRQS + ㄥRSQ = 180°  100° = 80°
ㄥRQS = ㄥRSQ = 40°
ㄥPQS = 180° – 40° = 140° (sum of angles on a line = 180°)
Then again, ㄥQPS = ㄥQSP (since angles opposite to equal sides are equal)
ㄥQPS + ㄥQSP = 180° – 140° = 40°
ㄥQPS = ㄥQSP = 20°
In the given figure, AD is the bisector of ΔBAC, AB = 6 cm, AC = 5 cm and BD = 3 cm. Find DC. It is given that ∠ABD = ∠ACD.
We know that the internal bisector of angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle.
Hence:
AB/BD = AC/CD
6/3 = 5/CD
CD = 2.5 cm
In the given figure, straight line AB and CD intersect at O. IF ∠δ =3∠v, then ∠v = ?
COD is a straight line
∴ ∠δ + ∠v =180⁰ ⇒ 3v +v =180 ⇒ 4v = 180 ⇒ v =45⁰.
In a triangle ABC, the incentre is at 0. If ㄥBOC = 100°, find ㄥBAC.
In ∠BOC
⇒ x + y = 80°
⇒ 2x + 2y. = 160°
Also, 2x + 2y + 2z = 180°
⇒ 160° + 2z = 180°
⇒ ∠BAC = 2z = 20°
A rectangular enclosure 40 m x 36 m has a horse tethered to a corner with a rope of 14 m in length. What is the ratio of the respective areas it can graze if it is outside the enclosure and if it is inside the enclosure?
Read the passage below and solve the questions based on it.
The area of a square is equal to the area of a rectangle. Moreover, the perimeter of the square is also equal to the perimeter of the rectangle.
Q. The length of the rectangle is equal to the:
This is possible only when both the length and breadth of the rectangle are equal to the side of the square.
A quadrilateral is inscribed in a circle. If an angle is inscribed in each of the four segments outside the quadrilateral, then what is the sum of these four angles?
Required sum = θ + 180°  θ + 180°  θ + 180°  θ + θ + 8 = 540°
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