Solve the problems below. answer with completely simplified exact value(s) or expression(s). given: δtsp, ts = sp = 10cm, tp = 12cm. find: three altitudes of δtsp.

The altitude to T is 9.6 cm

The altitude to S is 8 cm

The altitude to P is 9.6 cm

∆TSP is isosceles, so the altitude to TP from S will bisect TP (at point A) and create two right triangles, each with a base of (12 cm)/2 = 6 cm and a hypotenuse of 10 cm. The Pythagorean theorem tells you ...

... SP² = SA² +AP²

... (10 cm)² = SA² + (6 cm)²

... (100 -36) cm² = SA² = 64 cm²

... SA = √(64 cm²) = 8 cm . . . . the altitude to S

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The altitude to the other sides can now be found from the area of the triangle. The area of ∆TSP is ...

... A = (1/2)bh = (1/2)·TP·SA

... A = (1/2)(12 cm)(8 cm) = 48 cm²

We can call the point on SP where the altitude from T intersects it point B. Then the area is the same using base SP and altitude TB.

... A = (1/2)bh = (1/2)·SP·TB

... 48 cm² = (1/2)(10 cm)·TB

... (48 cm²)/(5 cm) = TB = 9.6 cm . . . . the altitude to T

The altitude from P to TS, the other side of length 10 cm, will be the same. The altitude to P is 9.6 cm.

look at the picture i have filled out a table for it. hope this .