SSC (English Medium) 10th Standard - Maharashtra State Board Question Bank Solutions for Geometry Mathematics 2

A milk container of height 16 cm is made of metal sheet in the form of frustum of a cone with radii of its lower and upper ends as 8 cm and 20 cm respectively. Find the cost of milk at the rate of ₹ 22 per litre which the container can hold.

Find the total surface area of frustum, if its radii are 15 cm and 7 cm. Also, the slant height of the frustum is 14 cm.

Radii of the frustum = `square` cm and `square` cm

Slant height of the frustum = `square` cm

Total surface area = `π[(r_1^2 + r_2^2 + (r_1 + r_2)l]`

= `22/7 [square + square + (square + square) square]`

= `22/7 (square)`

= `square` cm²

Hence, the total surface area of the frustum is `square`.

Draw a line segment AB of length 10 cm and divide it internally in the ratio of 2:5 Justify the division of line segment AB.

In the following figure, a quadrilateral LMNO circumscribes a circle with centre C. ∠O = 90°, LM = 25 cm, LO = 27 cm and MJ = 6 cm. Calculate the radius of the circle.

A pizza has 8 slices all equally spaced. Suppose pizza is a flat circle of radius 28 cm, find the area covered between 3 slices of pizza.

If radius of the base of cone is 7 cm and height is 24 cm, then find its slant height.

In ∆PQR seg PM is a median. Angle bisectors of ∠PMQ and ∠PMR intersect side PQ and side PR in points X and Y respectively. Prove that XY || QR. 

Complete the proof by filling in the boxes.

solution:

In ∆PMQ,

Ray MX is the bisector of ∠PMQ.

∴ `("MP")/("MQ") = square/square` .............(I) [Theorem of angle bisector]

Similarly, in ∆PMR, Ray MY is the bisector of ∠PMR.

∴ `("MP")/("MR") = square/square` .............(II) [Theorem of angle bisector]

But `("MP")/("MQ") = ("MP")/("MR")`  .............(III) [As M is the midpoint of QR.] 

Hence MQ = MR

∴ `("PX")/square = square/("YR")`  .............[From (I), (II) and (III)]

∴ XY || QR   .............[Converse of basic proportionality theorem]

In ΔABC, ray BD bisects ∠ABC, A – D – C, seg DE || side BC, A – E – B, then for showing `("AB")/("BC") = ("AE")/("EB")`, complete the following activity:

Proof :

In ΔABC, ray BD bisects ∠B.

∴ `square/("BC") = ("AD")/("DC")`   ...(I) (`square`)

ΔABC, DE || BC

∴ `(square)/("EB") = ("AD")/("DC")`   ...(II) (`square`)

∴ `("AB")/square = square/("EB")`   ...[from (I) and (II)]

In the following figure, AE = EF = AF = BE = CF = a, AT ⊥ BC. Show that AB = AC =  `sqrt3xxa`

In ∆PQR, PM = 15, PQ = 25 PR = 20, NR = 8. State whether line NM is parallel to side RQ. Give reason. 

In trapezium ABCD, side AB || side PQ || side DC, AP = 15, PD = 12, QC = 14, Find BQ. 

In the given figure, seg PA, seg QB, seg RC, and seg SD are perpendicular to line AD.

AB = 60, BC = 70, CD = 80, PS = 280 then find PQ, QR, and RS. 

In the given fig, XY || seg AC. If 2AX = 3BX and XY = 9. Complete the activity to Find the value of AC.  

Activity: 2AX = 3BX  

∴ `"AX"/"BX" = square/square`

`("AX" +"BX")/"BX" = (square + square)/square`   ...(by componendo)

`"AB"/"BX" = square/square`           ...(I)

ΔBCA ~ ΔBYX             ...`square` test of similarity,

∴ `"BA"/"BX" = "AC"/"XY"`    ...(corresponding sides of similar triangles)

∴ `square/square = "AC"/9`      

∴ AC = `square`        ...[From(I)]

In the given figure, the vertices of square DEFG are on the sides of ∆ABC. ∠A = 90°. Then prove that DE² = BD × EC. (Hint: Show that ∆GBD is similar to ∆CFE. Use GD = FE = DE.) 

Find the length of the side and perimeter of an equilateral triangle whose height is `sqrt3` cm.

∆ABC is an equilateral triangle. Point P is on base BC such that PC = `1/3`BC, if AB = 6 cm find AP.

From the information given in the figure, prove that PM = PN =  \[\sqrt{3}\]  × a

Find the length of the hypotenuse in a right angled triangle where the sum
of the squares of the sides making right angle is 169.
(A)15 (B) 13 (C) 5 (D) 12

Prove that, in a right angled triangle, the square of the hypotenuse is
equal to the sum of the squares of remaining two sides.

In right angled triangle PQR,
if ∠ Q = 90°, PR = 5,
QR = 4 then find PQ and hence find tan R.