Question for Practice Questions Level 2: Geometry - 1
Try yourself:If the medians of a right-angled triangle, which are drawn from the vertices of the acute angles, measure 6 cm and 8 cm, find the length of the hypotenuse.
Explanation
In ΔABC, AE and CD are the medians of the triangle.
So, AE = 6 and CD = 8
hence, BE = EC = 1/2BC and AD = DB= 1/2AB
Adding (l) and (2), we get
Question for Practice Questions Level 2: Geometry - 1
Try yourself:If O is the centre of the circle and R is the point of contact of the tangent, then what is the measure of ∠ORQ? Explanation
∠PQR = 50° (Alternate Segment Theorem)
∠PQO + ∠ OQR = 50°
30° + ∠OQR = 50°
∠OQR = 20°
Since PO = OQ, (Radii of the same circle)
∠OPQ = ∠OQP = 30°
∠OQR = ∠ORQ = 20°
Question for Practice Questions Level 2: Geometry - 1
Try yourself:Find the measure of angle PQS, if the measure of arc PMQ is 150°. Explanation
As shown in the figure, join PO and QO, where O is the centre.
Arc PMQ = 150°
(∵ Given ∠POQ = 150°)
Now, join PN and QN as shown in the figure.
∠PNQ = 1/2 ∠POQ = 75° [∵ The angle subtended by an arc or a chord of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle]
Now, ∠PQS = ∠PNQ [∵ Angles in alternate segments are always equal]
∠PQS = 75°
Question for Practice Questions Level 2: Geometry - 1
Try yourself:What is the length of PN, if AB = 30, PM = 8, DC = 16 and P is the centre of the circle? Explanation
PM ⊥ AB
Any perpendicular from the centre to the chord bisects the chord.
In ΔAMP
AP^{2} = MP^{2} + AM^{2}
AP^{2} = 82 + 152 = 64 + 225 = 289
In ΔCNP
CP^{2} = NP^{2} + CN^{2}
17^{2} = NP^{2} + 8^{2}
17^{2} - 8^{2} = NP^{2}
NP = 15
Question for Practice Questions Level 2: Geometry - 1
Try yourself:Four tangents to a circle form a parallelogram PQRS. Find the length of side PS. Explanation
Opposite sides of a parallelogram are parallel and equal.
Therefore, PQ = SR and PS = QR.
Suppose the length of SC is x units and that of CR is y units.
∵ SR = PQ
∴ x + y = 8 ... (1)
SC and SD are tangents to the circle, so SC = SD = x units.
CR and RB are tangents to the circle, so CR = BR = y units.
SP = (x + 3) = RQ = (5 + y)
x - y = 2 ... (2)
Solving the equations (1) and (2), we get
x = 5 units and y = 3 units
∴ PS = 3 + x = 3 + 5 = 8 units
Question for Practice Questions Level 2: Geometry - 1
Try yourself:If PS || QR, ∠1 = ∠2 and ∠3 = ∠4, then what is the measure of ∠STR? Explanation
Given; PS || RQ, ∠1 = ∠2, ∠3 = ∠4
Let ∠STR = θ
∠PSR + ∠SRQ = 180° {Sum of co-interior angles is 180°}
∠1 + ∠2 + ∠3 + ∠4 = 180°
2∠2 + 2∠3 = 180°
∠2 + ∠3 = 90° ...(1)
Now, ∠2 + ∠θ + ∠3 = 180° {Sum of the angles in a triangle is 180°}
∠θ + 90° = 180° {from (1)}
∠θ = 180° - 90°
∠θ = 90°
∴ ∠STR = ∠θ = 90°
Question for Practice Questions Level 2: Geometry - 1
Try yourself:The sum of the interior angles of a triangle is 180°. How many sides does a polygon have if the sum of its interior angles is 2520°?
Explanation
Sum of interior angles of a polygon = (n - 2) x 180°, where n is the number of sides of polygon.
Now, (n - 2) x 180° = 2520°
n - 2 = 14
n = 16
Question for Practice Questions Level 2: Geometry - 1
Try yourself:In the given figure, AB || CD. Find angle BGF. Explanation
Given, AB || CD.
70° = 30° +∠ECD
∠ECD = 40°
∠ECD + ∠CEF = 140° + 40° = 180°
So, EF || CD.
∠EFG = ∠FGB
∠BGF = 50°
Question for Practice Questions Level 2: Geometry - 1
Try yourself:In the following figure, OA bisects ∠A, ∠ABO = ∠ACO and ∠BOC = 100°. Find the measure of ∠AOB. Explanation
In triangle AOB and AOC, two angles are correspondingly equal.
⇒ ∠AOB = ∠AOC = z
Now, ∠AOB + ∠AOC + ∠BOC = 360°
z + z + 100° = 360°
Or z = 130°
Hence, option (2) is correct.
Question for Practice Questions Level 2: Geometry - 1
Try yourself:Six circles, each of radius r feet, are placed as shown in the figure below. If A is the area bounded by but not included in the circles (in square feet), and B is the height of the stack (in feet), then find the value of A + Br. Explanation
Draw a triangle running through the centres of the circles as shown above. Since the circles are of equal radii, ΔABC will be an equilateral triangle with side 4r.
Area bounded by but not included in the circles (A) = Area of the equilateral triangle – (Area of 3 semicircles + Area of 3 sectors of angle 60° each)