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The diagram represents the cross-sections of a loft PQRST, PTQ is an isosceles triangle and QRST is a rectangle. The height PN of P above the ground is 7.5 m. The height QR is 5 m and PQ is 6.5 m. - Mathematics

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Question

The diagram represents the cross-sections of a loft PQRST, PTQ is an isosceles triangle and QRST is a rectangle.


The height PN of P above the ground is 7.5 m. The height QR is 5 m and PQ is 6.5 m. Given that N is the mid-point of SR, find the length of SR and the area of PQRST.

Sum
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Solution

Given:

  • PQRST is a loft where PTQ forms an isosceles triangle and QRST forms a rectangle.
  • Height PN height of point P above ground = 7.5 m
  • Height QR height of rectangle QRST = 5 m
  • PQ = 6.5 m
  • N is the midpoint of SR.

We need to find:

  1. The length of SR
  2. The area of the entire loft PQRST.

Step 1: Understanding the figure and labeling

Since QRST is a rectangle and QR = 5 m height, then SR is the base of the rectangle and lies on the ground.

N is the midpoint of SR, so N divides SR into two equal segments.

PTQ is an isosceles triangle with PQ = PT = 6.5 m.

The height PN from P to line SR is 7.5 m.

Since N is midpoint of SR, triangles PNT and PNQ are right triangles.

Step 2: Find the half length of SR using right triangle PNT or PNQ

In triangle PNQ,

PN = 7.5 m   ...(Height)

PQ = 6.5 m   ...(Equal side)

Use Pythagoras theorem to find NQ half of SR:

PQ2 = PN2 + NQ2 

(6.5)2 = (7.5)2 + NQ2

42.25 = 56.25 + NQ2

NQ2 = 42.25 – 56.25

NQ2 = –14   ...(Which is not possible)

There is a mistake here; since the hypotenuse PQ = 6.5 is less than height PN = 7.5 m, this is contradictory.

Reviewing given data typically, in an isosceles triangle with vertex P and base TQ on the ground, the height from P to TQ should be less than side PQ.

Therefore, reassess the triangle is PQ the side or the base?

Given PTQ is isosceles triangle with equal sides PT and PQ = 6.5 m each and base TQ unknown.

Assuming N is midpoint of SR and QRST is rectangle with height QR = 5 m, so SR is horizontal base line.

Since N is midpoint of SR and PN = 7.5 m is height from point P to SR, the vertical distance between P and SR is 7.5 m.

Point Q is directly above R with height 5 m; thus, Q lies 5 m above R.

So point Q has a height of 5 m and point P has height 7.5 m, meaning vertical distance between P and Q in same horizontal position vertically is 2.5 m.

Step 3: Determine length of NQ = ?

From PQ = 6.5 m, in triangle PNQ:

PN = 7.5 m

PQ = 6.5 m

NQ = ?

Use Pythagoras:

PQ2 = PN2 + NQ2

6.52 = 7.52 + NQ2

42.25 = 56.25 + NQ2

NQ2 = 42.25 – 56.25

NQ2 = –14

Negative value implies triangle configuration not right angled with PN perpendicular to SR at N.

Instead, from the figure, PN is perpendicular to SR, so PN ⊥ SR. So in triangle PNS, PN is height.

So we must consider a right triangle with hypotenuse PT or PQ.

Recognize that since PTQ is isosceles with PT = PQ, then base TQ can be found.

Step 4: Use triangle PTQ properties to find base TQ

By the properties of an isosceles triangle with vertex P and PT = PQ = 6.5 m, height PN = 7.5 m from P to SR midpoint N of base SR, we use the distance from N to T or Q to find half the base TQ.

Since N is midpoint of SR and QRST is rectangle with height 5 and base SR to be found.

Step 5: Calculate half base TQ using triangle PNT right triangle

In the right triangle PNT,

PN = 7.5 m   ...(Height)

PT = 6.5 m   ...(Hypotenuse)

Find NT half TQ:

NT2 = PT2 – PN2

= 6.52 – 7.52

= 42.25 – 56.25

= –14   ...(Not possible)

This reinforces that height PN = 7.5 m is higher than side PT = 6.5 m, causing contradiction.

Step 6: Review problem assumptions and data

Since height PN (7.5 m) is greater than PQ or PT (6.5 m), this is impossible for a triangle.

Alternatively, perhaps PN is the height of P above the ground (SR) and height QR = 5 m is height of rectangle, so point Q lies 5 m above R.

Since N is midpoint of SR, and height from P to N is 7.5 m, the vertical difference P and Q is 7.5 – 5 = 2.5 m.

So, vertical projection of P to Q is 2.5 m.

Distance between P and Q is 6.5 m.

Let the horizontal distance between N and Q be x.

Use Pythagoras in triangle PNQ:

PQ2 = PNQ2 + NQ2

6.52 = (2.5)2 + x2

42.25 = 6.25 + x²

x2 = 36

x = 6 m

So NQ = 6 m

Step 7: Since N is midpoint of SR and QRST is rectangle

SR = 2 × NQ

= 2 × 6

= 12 m

Step 8: Calculate area of PQRST

Area of rectangle QRST = base SR × height QR

= 12 m × 5 m

= 60 m2

Area of triangle PTQ = 0.5 × base TQ × height PN above base TQ

Length of TQ = 2 × NQ

= 2 × 6

= 12 m

Height of triangle P above TQ is PN – QR

Height = 7.5 m – 5 m

Height = 2.5 m? No.

Because TQ is at height 5 m, P is at 7.5 m.

So height of triangle PTQ above base TQ is P height – Q height

= 7.5 – 5

= 2.5 m

So area of triangle PTQ

= 0.5 × base TQ × height 

= 0.5 × 12 × 2.5

= 15 m2

Step 9: Total area of PQRST

= Area of rectangle QRST + Area of triangle PTQ

= 60 m2 + 15 m2

= 75 m2

Length of SR = 12 m

Area of PQRST = 75 m2

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Chapter 17: Mensuration - EXERCISE 17B [Page 206]

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B Nirmala Shastry Mathematics [English] Class 9 ICSE
Chapter 17 Mensuration
EXERCISE 17B | Q 13. | Page 206
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