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The area of triangle LMN is 120 square units. The shaded and unshaded regions have equal areas. If the area of triangle MVU is 1/3rd the area of triangle LMN, and the area of triangle LWX is 1/4^{th} the area of triangle LMN, then find the area of quadrilateral ABCD. (Note: The figure is not drawn as per the scale.)
Area of ΔMUV = (1/3) x Area of ΔLMN = 40 sq. units
Similarly, area of ΔLWX = (1/4) x 120 sq. units = 30 sq. units
Let the area of ABCD be x sq. units.
Then, area of the shaded region = 30 sq. units – x + 40 sq. units – x + x = 70 sq. units – x = Area of the unshaded region
∴ 70 sq. units – x = (½) × 120 sq. units = 60 sq.units
∴ x = 10 sq. units
A wheel rests against the edge of a pavement as shown below. The height of the pavement is 3 m and the distance of the point where the wheel touches the ground from the point where the wheel touches the edge of the pavement is 5 m. Find the radius of the wheel.
Draw BD  AC to cut OA at D.
In triangle ABC,
3^{2} + AC^{2} = 5^{2}
AC = 4 m
Let radius = r.
OD^{2} + DB^{2} = OB^{2}
DB = AC
DB^{2} = r^{2}  (r  a)^{2}
4^{2} = 2ar  a^{2 } (where a = 3)
16 + 9 = 2 x 3 x r
or, r = (25/6) m
A circle is inscribed in a quadrant of a bigger circle, as shown in the figure below.
If the radius of the smaller circle is 2 units, what is the radius of the bigger circle?
PQ = PS = 2
⇒ OQ = 2 [∵ ΔOQP ≅ ΔOSP]
Let OA = r
OP = (OC  PC) = r  2
Also, (OP)^{2} = 2(OQ)^{2}
(r  2)^{2} = 2(2)^{2}
⇒ r^{2} + 4  4r = 8
⇒ r^{2}  4r  4 = 0
⇒ r = 2(1 +√2) = 2(√2 + 1)
As shown in the figure, PQRS is a square and TQUV is another square inside it in a corner. VQ = QS. Find the ratio of the area of ΔVSU to the area of PQRS.
The diagrammatic representation for the same is as shown
Thus, answer option 3 is correct
In the figure (not drawn to scale), rectangle ABCD is inscribed in a circle with centre at O. The length of side AB is greater than that of side BC. The ratio of the area of the circle to the area of the rectangle ABCD is π :. The line segment DE intersects AB at E such that ∠ODC = ∠ADE.
What is the ratio of AE : AD?
Let the radius of the circle be 'R' and ∠ODC = ∠ADE = θ.
If OM is drawn perpendicular to DC,
DM = R cos θ
OM = R sin θ
Length of rectangle ABCD,
AB = CD = 2DM = 2R cos θ
AD = BC = 2OM = 2R sin θ
Area of rectangle = AB(BC) = 2R cos θ x 2R sin θ = 2R² sin2θ
Area of circle = πR²
According to the question,
πR² : 2R² sin 2θ = π : √3
⇒ sin 2θ = √2 = sin 60°
θ = 30°
In triangle ADE, tan θ = AE/AD
AE : AD = tan 30° = 1 : √3
In the figure, the rectangle at the corner measures 10 cm x 20 cm. One corner of the rectangle is also a point on the circumference of the circle. What is the radius of the circle?
Let the radius of the circle be `r` cm.
∴ We get (r  20)^{2} + (r  10)^{2} = r^{2}
∴ r^{2}  40r + 400 + r^{2}  20r + 100 = r^{2}
∴ r^{2}  60r + 500 = 0
∴ r = 10 cm or r = 50 cm
But, r cannot be 10 cm.
∴ r = 50 cm
The area of an equilateral triangle DRS is 144√3 sq. cm. If P is the centre of the circle, find the area of quadrilateral DRPS in the given figure.
As shown in the figure, area of equilateral triangle
(a represents the length of side of an equilateral triangle)
a = 24 cm
In triangle RDP,
∠RDP = 30°
cos 30° = RD/PD = 24/PD
PD = 16√3 cm
Area of quadrilateral RDSP
192√3 sq. cm^{2}
As shown in the figure, Area of quadrilateral PQRS = Area of triangle PQR + Area of triangle PRS … (1)
Both these triangles are rightangled triangles.
Area of triangle PQR = (1/2) x 4 x 3 = 6 sq. units
Area of triangle PRS = (1/2) x 12 x 5 = 30 sq. units
Area of quadrilateral PQRS = 6 + 30 = 36 sq. units
A, B, C, D, E ........ Z are the points marked on the circumference of a circle equidistantly. What can be the maximum number of triangles which can be formed using three points as vertices such that their circumcentre lies on one of the sides of a triangle?
Circumcentre will lie on one of the sides if the triangle is a rightangled triangle.
So, two of the vertices of the triangle will lie on a diameter of the circle.
So, the points must be equidistant to maximise the number of such triangles.
Let AN be one such chosen diameter and hence, one side of a triangle.
It is to be noted that there will be 13 such diameters possible.
So, third vertex of the triangle can be chosen from any of the other 24 points left (other 24 points leaving A and N).
Possible total number of such triangles = 13 × 24 = 312.
In the given figure, O is the centre of the circle and AE is a diameter. If AB = BC and ∠BFC = 25°, find the value of ∠ABC.
Angles subtended by equal chords are equal.
∠AFB = 25° = ∠BFC
In the cyclic quadrilateral ABCF,
∠AFC = 25° + 25° = 50°
∠ABC = 180°  50° = 130° (because the sum of opposite angles of a cyclic quadrilateral is 180°)
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