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rodikova [14]
4 years ago
7

Given: ΔTSP, TS = SP = 10cm, TP = 12cm. Find: Three altitudes of ΔTSP.

Mathematics
1 answer:
fredd [130]4 years ago
4 0

Answer:

The altitudes of ΔTSP is SC is 8 unit ,

PA is 5\sqrt{3} unit ,

TB is 5\sqrt{3} unit .

Step-by-step explanation:

Given as :

The Triangle Δ TSP having side TS , SP , TP

The measure of TS = 12 cm

The measure of SP = 10 cm

The measure of TP = 10 cm

Let The altitude from point T on side PS = TB

The altitude from point S on side PT = SC

The altitude from point P on side TS = PA

The altitude divide the sides

TS = TA + AS

PS = PB + PS

PT = PC + CT

So, From Pythagoras theorem

PA² = PS² - AS²

PA² = 10² - 5²

Or, PA² = 100 - 25

Or, PA = \sqrt{75}

∴  PA = 5\sqrt{3} unit

Again

TB² = TS² - BS²

TB² = 10² - 5²

Or, TB² = 100 - 25

Or, TB = \sqrt{75}

∴ TB =  5\sqrt{3} unit

Similarly

SC² = TS² - TC²

SC² = 10² - 6²

Or, SC² = 100 - 36

Or, SC = \sqrt{64}

∴ SC = 8 unit

Hence The altitudes of ΔTSP is SC is 8 unit , PA is 5\sqrt{3} unit , TB is 5\sqrt{3} unit . Answer

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